Note that Exponential and Logarithmic Differentiation is covered here.
Exponential and Logarithmic Differentiation and Integration have a lot of practical applications and are handled a little differently than we are used to. For a review of these functions, visit the Exponential Functions section and the Logarithmic Functions section.
Introduction to Exponential and Logarithmic Differentiation and Integration
Before getting started, here is a table of the most common Exponential and Logarithmic formulas for Differentiation and Integration:
| Exponential and Logarithmic Derivatives | Exponential and Logarithmic Integrals | Examples |
| $ \displaystyle \frac{d}{{dx}}\left( {\ln u} \right)=\frac{{{u}’}}{u}$ | $ \displaystyle \int{{\frac{{{u}’}}{u}}}\,dx=\ln \left| u \right|\,+C$ (We need absolute value since $ \ln u$ is only defined for $ u>0$ .) | $ \displaystyle \color{#800000}{{\frac{{d\left[ {\ln \left( {{{x}^{5}}-3} \right)} \right]}}{{dx}}}}=\frac{{{u}’}}{u}=\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}\,\,\,\,\,\,\,\text{(}u={{x}^{5}}-3)$
$ \displaystyle \begin{align}u&={{x}^{5}}-3\\du&=5{{x}^{4}}\,dx\\dx&=\frac{{du}}{{5{{x}^{4}}}}\end{align}$ $ \require{cancel} \begin{align} \color{#800000}{{\int{{\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}}}\,dx}}&=\int{{\frac{{5{{x}^{4}}}}{u}}}\,dx=\int{{\frac{{\cancel{{5{{x}^{4}}}}}}{u}}}\,\cdot \frac{{du}}{{\cancel{{5{{x}^{4}}}}}}\\&=\int{{\frac{{du}}{u}}}\,=\ln \left| u \right|+C=\ln \left| {{{x}^{5}}-3} \right|+C\end{align}$
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| $ \displaystyle \frac{d}{{dx}}\left( {{{{\log }}_{a}}u} \right)=\frac{{{u}’}}{{u\left( {\ln \,a} \right)}}$ | $ \displaystyle \int{{\frac{{{u}’}}{{u\left( {\ln a} \right)}}}}\,dx={{\log }_{a}}\left| u \right|\,+C$ (Not a typical integration problem) | $ \displaystyle \color{#800000}{{\frac{d}{{dx}}\left[ {\log \left( {4x-1} \right)} \right]}}=\frac{d}{{dx}}\left[ {{{{\log }}_{{10}}}\left( {4x-1} \right)} \right]=\frac{4}{{\left( {4x-1} \right)\ln 10}}$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{e}^{u}}} \right)={{e}^{u}}{u}’$ | $ \displaystyle \int{{{{e}^{u}}}}={{e}^{u}}+C$ | $ \displaystyle \color{#800000}{{\frac{{d\left( {{{e}^{{3x}}}} \right)}}{{dx}}}}={{e}^{{3x}}}\cdot 3=3{{e}^{{3x}}}$ $ \begin{align}u&=3x\\du&=3\,dx\\dx&=\frac{{du}}{3}\end{align}$ $ \displaystyle \color{#800000}{{\int{{3{{e}^{{3x}}}}}dx}}=3\int{{{{e}^{u}}dx}}=\cancel{3}\int{{{{e}^{u}}\cdot \frac{{du}}{{\cancel{3}}}}}={{e}^{u}}+C={{e}^{{3x}}}+C$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{a}^{u}}} \right)=\left( {\ln \,a} \right){{a}^{u}}{u}’$ | $ \displaystyle \int{{{{a}^{u}}}}du=\left( {\frac{1}{{\ln a}}} \right){{a}^{u}}+C$ | $ \displaystyle \color{#800000}{{\frac{d}{{dx}}\left( {{{4}^{{3{{x}^{3}}}}}} \right)}}=\ln 4\left( {{{4}^{{3{{x}^{3}}}}}} \right)\left( {9{{x}^{2}}} \right)$
$ \displaystyle \begin{align}u&=3{{x}^{3}}\\du&=9{{x}^{2}}\,dx\\dx&=\frac{{du}}{{9{{x}^{2}}}}\end{align}$ $ \displaystyle \begin{align}\color{#800000}{{\int{{\ln 4\left( {{{4}^{{3{{x}^{3}}}}}} \right)}}\left( {9{{x}^{2}}} \right)dx}}&=\ln 4\int{{\left( {{{4}^{u}}} \right)}}\left( {\cancel{{9{{x}^{2}}}}} \right)\frac{{du}}{{\cancel{{9{{x}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,&=\cancel{{\ln 4}}\cdot \frac{1}{{\cancel{{\ln 4}}}}\cdot {{4}^{u}}+C={{4}^{{3{{x}^{3}}}}}+C\end{align}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {f{{{\left( x \right)}}^{{g\left( x \right)}}}} \right]$ When we have a variable both in the base and the exponent, or if the function is really complicated, it’s best to take ln of both sides to take derivative, and use implicit integration. Then substitute $ y$ back in at the end. | Find Derivative: $ \displaystyle y={{\left( {3x} \right)}^{x}}$ $ \displaystyle \begin{align}\ln y&=\ln {{\left( {3x} \right)}^{x}};\,\,\,\,\,\ln y=x\ln \left( {3x} \right)\\\frac{1}{y}\left( {\frac{{dy}}{{dx}}} \right)&=\cancel{x}\cdot \frac{{\cancel{3}}}{{\cancel{3}\cancel{x}}}+\ln \left( {3x} \right)\cdot 1\,\,\,\,\text{(product rule)}\\\,\frac{{dy}}{{dx}}&=\left[ {1+\ln \left( {3x} \right)} \right]\cdot y={{\left( {3x} \right)}^{x}}\left[ {1+\ln \left( {3x} \right)} \right]={{3}^{x}}{{x}^{x}}\left[ {1+\ln \left( {3x} \right)} \right]\end{align}$ | |
| Integrals of the Six Basic Trig Functions $ \displaystyle \int{{\sin u\,du=-\cos u\,\,+\,\,C\,\,\,\,\,\,\,\,}}\int{{\cos u\,du=\sin u\,+\,C}}$ $ \displaystyle \int{{\tan u\,du=-\ln \left| {\cos u} \right|\,\,+\,\,C\,}}\,\,\,\,\,\,\,\,\int{{\cot u\,du=\ln \left| {\sin u} \right|\,+\,C}}$ $ \displaystyle \int{{\sec u\,du=\ln \left| {\sec u+\tan u} \right|\,\,+\,\,C}}\,\,\,\,\,\,\,\int{{\csc u\,du=-\ln \left| {\csc u+\cot u} \right|\,+\,C}}$ | $ \begin{align}u&=4\theta \\du&=4\,d\theta \\d\theta &=\frac{{du}}{4}\end{align}$ $ \displaystyle \begin{align}\color{#800000}{{\int{{\csc \left( {4\theta } \right)}}\,d\theta }}&=\int{{\csc u}}\,d\theta =\int{{\csc u\cdot \,\frac{{du}}{4}}}=\frac{1}{4}\int{{\csc u\,du}}\\&=-\frac{1}{4}\ln \left| {\csc u+\cot u} \right|\,+\,C\\&=-\frac{1}{4}\ln \left| {\csc \left( {4x} \right)+\cot \left( {4x} \right)} \right|\,+\,C\end{align}$ | |
Actually, when we take the integrals of exponential and logarithmic functions, we’ll be using a lot of U-Substitution Integration, so you may want to review it.
Review of Logarithms
When we learned the Power Rule for Integration here in the Antiderivatives and Integration section, we noticed that if $ n=-1$, the rule doesn’t apply since the denominator would be $ 0$: $ \displaystyle \int{{{{x}^{n}}}}dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}\,+C,\,\,n\ne 1$. So, when we try to integrate a function like $ \displaystyle f\left( x \right)=\frac{1}{x}={{x}^{{-1}}}$, we have to do something “special”; namely learn that this integral is $ \ln \left( x \right)$. Remember that $ \ln x$ is the same as $ {{\log }_{e}}x$, where $ e\approx 2.718$ (“$ e$” is Euler’s Number). A log is the exponent raised to the base power ($ a$) to get the argument ($ x$) of the log (if “$ a$” is missing, we assume it’s 10).
Here are some logarithmic properties that we learned here in the Logarithmic Functions section; note we could use $ {{\log }_{a}}x$ instead of $ \ln x$.
Logarithmic Properties
$ \begin{array}{c}\ln \left( {{{a}^{n}}} \right)=n\ln a\\\ln \left( {ab} \right)=\ln \left( a \right)+\ln \left( b \right)\\\ln \left( {\frac{a}{b}} \right)=\ln \left( a \right)-\ln \left( b \right)\\\ln \left( 1 \right)=0\\\ln \left( {{{e}^{x}}} \right)=x\end{array}$ $ \displaystyle \begin{align}&\text{Example of Expanding Log:}\\\ln {{\left( {\frac{{{{x}^{4}}y}}{{\sqrt{z}}}} \right)}^{2}}&=2\ln \left( {\frac{{{{x}^{4}}y}}{{{{z}^{{\frac{1}{2}}}}}}} \right)=2\left[ {\ln \left( {{{x}^{4}}} \right)+\ln \left( y \right)-\ln \left( {{{z}^{{\frac{1}{2}}}}} \right)} \right]\\&=2\left[ {4\ln \left( x \right)+\ln \left( y \right)-\frac{1}{2}\ln \left( z \right)} \right]\\&=8\ln \left( x \right)+2\ln \left( y \right)-\ln \left( z \right)\end{align}$
The Log Rule for Integration
We learned that the differentiation rule for log functions is $ \displaystyle \frac{d}{{dx}}\left[ {\ln u} \right]du=\frac{{{u}’}}{u}$.
From this, we can get the Log Rules for Integration; you’ll probably just want to memorize these. Remember that | | is the Absolute Value Function, which means always take the positive of what’s inside. The reason we use an absolute value is that the natural logarithm function is only defined for $ x>0$.
Log Rules for Integration
$ \displaystyle \int{{\frac{1}{x}}}\,dx=\ln \left| x \right|\,+C$ $ \displaystyle \int{{\frac{{{u}’}}{u}}}\,dx=\ln \left| u \right|\,+C$
Most of these problems involve U-Substitution Integration and some require doing Polynomial Long Division before integrating when the degree (largest exponent of all the terms) of the numerator is greater than the degree of the denominator.
Here are some Logarithmic Integration problems; assume that $ x$ cannot be equal to any value that makes a denominator 0:
| Log Integration Problem | Solution |
| $ \displaystyle \int{{\frac{2}{{x+5}}}}\,dx$ | $ \displaystyle \int{{\frac{2}{{x+5}}}}\,dx=2\int{{\frac{1}{{x+5}}}}\,dx=2\ln \left| {x+5} \right|+C$ |
| $ \displaystyle \int{{\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}}}\,dx$ | $ \begin{align}u&={{x}^{5}}-3\\du&=5{{x}^{4}}\,dx\end{align}$ $ \displaystyle \begin{align}\int{{\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}}}\,dx&=\int{{\frac{{du}}{u}}}\,\,\left( {\text{or }\int{{\frac{1}{u}du}}} \right)\\&=\ln \left| u \right|+C=\ln \left| {{{x}^{5}}-3} \right|+C\end{align}$ |
| $ \displaystyle \int{{\frac{{{{x}^{2}}-8}}{x}}}\,dx$ | Degree on top larger: divide by $ x$ first: $ \displaystyle \int{{\frac{{{{x}^{2}}-8}}{x}}}\,dx=\int{{\left( {x-\frac{8}{x}} \right)}}\,dx=\int{{x\,dx-8\int{{\frac{1}{x}}}\,dx}}=\frac{1}{2}{{x}^{2}}-8\ln \left| x \right|+C$ |
| $ \displaystyle \int{{\frac{{{{{\left( {\ln x} \right)}}^{3}}}}{{4x}}}}\,dx$ (or $ \displaystyle \int{{\frac{{{{{\ln }}^{3}}\left( x \right)}}{{4x}}}}\,dx$) | $ \displaystyle \begin{align}u&=\ln x\\du&=\frac{1}{x}\,dx;\,\,\,dx=x\,du\end{align}$ $ \begin{align}\int{{\frac{{{{{\left( {\ln x} \right)}}^{3}}}}{{4x}}}}\,dx&=\int{{\frac{{{{u}^{3}}}}{{4\cancel{x}}}\cancel{x}\,du}}=\frac{1}{4}\int{{{{u}^{3}}\,du}}\\&=\frac{1}{4}\cdot \frac{{{{u}^{4}}}}{4}+C=\frac{{{{{\left( {\ln x} \right)}}^{4}}}}{{16}}+C\,\,\text{(or }\frac{{{{{\ln }}^{4}}\left( x \right)}}{{16}}+C)\end{align}$ |
| $ \displaystyle \int{{\frac{{{{x}^{3}}-4{{x}^{2}}+x-3}}{{x+2}}}}\,dx$ | Degree on top larger: Long Division:
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| Solve the Differential Equation passing through the point $ \left( {0,4} \right)$:
$ \displaystyle \frac{{dy}}{{dx}}=\frac{{3x}}{{{{x}^{2}}-1}}$ | $ \displaystyle \frac{{dy}}{{dx}}=\frac{{3x}}{{{{x}^{2}}-1}};\,\,\,\,\,dy=\frac{{3x}}{{{{x}^{2}}-1}}\,dx;\,\,\,\,\,\,\,y=\int{{\frac{{3x}}{{{{x}^{2}}-1}}dx}}$
$ \displaystyle \begin{align}u&={{x}^{2}}-1\\du&=2x\,dx;\,\,\,\,dx=\frac{{du}}{{2x}}\end{align}$ $ \begin{align}\int{{\frac{{3x}}{{{{x}^{2}}-1}}}}\,dx&=\int{{\frac{{3\cancel{x}}}{u}}}\cdot \frac{{du}}{{2\cancel{x}}}=\frac{3}{2}\int{{\frac{1}{u}\,du}}\\&=\frac{3}{2}\ln \left| u \right|+C=\frac{3}{2}\ln \left| {{{x}^{2}}-1} \right|+C\\y&=\frac{3}{2}\ln \left| {{{x}^{2}}-1} \right|+C\end{align}$ $ \begin{align}\text{Solve for C at point}(0,4):\,\,4&=\frac{3}{2}\ln \left| {{{{\left( 0 \right)}}^{2}}-1} \right|+C;\,\,\,\,C=4-\frac{3}{2}\ln \left( 1 \right)=4-0=4\\y&=\frac{3}{2}\ln \left| {{{x}^{2}}-1} \right|+4\end{align}$ |
| $ \displaystyle \int\limits_{1}^{e}{{\frac{{{{{\left( {2+\ln x} \right)}}^{2}}}}{{4x}}}}\,dx$ | $ \begin{align}u&=2+\ln x\\du&=\frac{1}{x}\,dx;\,\,\,dx=x\,du\end{align}$ $ \displaystyle \begin{align}\int\limits_{1}^{e}{{\frac{{{{{\left( {2+\ln x} \right)}}^{2}}}}{{4x}}}}\,dx&=\int\limits_{{x=1}}^{{x=e}}{{\frac{{{{u}^{2}}}}{{4\cancel{x}}}}}\,\cancel{x}\,du=\frac{1}{4}\int\limits_{{x=1}}^{{x=e}}{{{{u}^{2}}\,du}}\\&=\left[ {\frac{1}{{12}}{{u}^{3}}} \right]_{{x=1}}^{{x=e}}=\left[ {\frac{1}{{12}}{{{\left( {2+\ln x} \right)}}^{3}}} \right]_{1}^{e}=\frac{1}{{12}}\left[ {{{{\left( {2+\ln e} \right)}}^{3}}-{{{\left( {2+\ln 1} \right)}}^{3}}} \right]\\&=\frac{1}{{12}}\left[ {{{{\left( {2+1} \right)}}^{3}}-{{{\left( {2+0} \right)}}^{3}}} \right]=\frac{{19}}{{12}}\end{align}$ |
$ \displaystyle \begin{align}\,\,\,\,&\,\,\int{{\left( {{{x}^{2}}-6x+13-\frac{{29}}{{x+2}}} \right)}}\,dx\\\,&=\int{{{{x}^{2}}\,dx-\int{{6x}}\,dx+\int{{13}}\,dx-\int{{\frac{{29}}{{x+2}}}}}}\,dx\\&=\frac{1}{3}{{x}^{3}}-3{{x}^{2}}+13x-29\ln \left| {x+2} \right|+C\\\end{align}$Integrals of Trigonometric Functions using “ln”
We learned integrals of some of the trig functions here in the Antiderivatives and Indefinite Integration section (included below), but now that we know some log rules, we’ll introduce the rest of the trig integrals. The new trig integrals may be proved by using the log integration rules, but you’ll probably just want to memorize these:
Integrals of the Six Basic Trig Functions
$ \begin{align}\int{{\sin u\,du}}&=-\cos u+C\\\int{{\tan u\,du}}&=-\ln \left| {\cos u} \right|+C\\\int{{\sec u\,du}}&=\ln \left| {\sec u+\tan u} \right|+C\end{align}$ $ \begin{align}\int{{\cos u\,du}}&=\sin u+C\\\int{{\cot u\,du}}&=\ln \left| {\sin u} \right|+C\\\int{{\csc u\,du}}&=-\ln \left| {\csc u+\cot u} \right|+C\end{align}$ $ \begin{align}\text{Earlier}\,\text{learned:}\\\,\,\,\int{{{{{\sec }}^{2}}x\,dx}}&=\tan x+C\\\int{{\sec x\tan x\,dx}}&=\sec x+C\\\int{{{{{\csc }}^{2}}}}x\,dx&=-\cot x+C\\\int{{\csc x\cot x\,dx}}&=-\csc x+C\end{align}$
Here are some Trig Integration problems; notice that sometimes we can’t use the above equations, but have to work with the log integral rules:
| Trig Integration | Solution |
| $ \displaystyle \int{{\csc \left( {4\theta } \right)}}\,d\theta $ | $ \begin{align}u&=4\theta \\du&=4\,d\theta \end{align}$ $ \displaystyle \begin{align}\int{{\csc \left( {4\theta } \right)}}\,d\theta& =\frac{1}{4}\int{{\csc u\,du}}\\&=-\frac{1}{4}\ln \left| {\csc \left( {4x} \right)+\cot \left( {4x} \right)} \right|+C\end{align}$ |
| $ \displaystyle \int{{{{t}^{2}}\cot \left( {{{t}^{3}}} \right)}}\,dt$ | $ \displaystyle \begin{align}u={{t}^{3}}&\\du=3{{t}^{2}}\,dt;\,\,dt&=\frac{{du}}{{3{{t}^{2}}}}\end{align}$ $ \displaystyle \begin{align}\int{{{{t}^{2}}\cot \left( {{{t}^{3}}} \right)}}\,dt&=\int{{\cancel{{{{t}^{2}}}}\cot \left( u \right)}}\,\,\frac{{du}}{{3\cancel{{{{t}^{2}}}}}}=\frac{1}{3}\ln \left| {\sin u} \right|+C\\&=\frac{1}{3}\ln \left| {\sin {{t}^{3}}} \right|+C\end{align}$ |
| $ \displaystyle \int{{\frac{{\cos t}}{{2\sin t-1}}}}\,dt$ | $ \begin{align}u=2\sin t-1&\\du=2\cos t\,dt;\,\,dt&=\frac{{du}}{{2\cos t}}\end{align}$ $ \displaystyle \begin{align}\int{{\frac{{\cos t}}{{2\sin t-1}}}}\,dt&=\int{{\frac{{\cancel{{\cos t}}}}{u}\cdot }}\frac{{du}}{{2\cancel{{\cos t}}}}=\frac{1}{2}\int{{\frac{1}{u}du}}\\&=\frac{1}{2}\ln \left| {2\sin t-1} \right|+C\end{align}$ |
| $ \displaystyle \int{{\frac{{\sec \theta \tan \theta }}{{3\sec \theta -1}}}}\,d\theta $ | $ \begin{align}u&=3\sec \theta -1\\du&=3\sec \theta \,\text{tan}\theta \,d\theta \\d\theta &=\frac{{du}}{{\sec \theta \,\text{tan}\theta }}\end{align}$ $ \begin{align}\int{{\frac{{\sec \theta \tan \theta }}{{3\sec \theta -1}}}}\,d\theta &=\int{{\frac{{\cancel{{\sec \theta \tan \theta }}}}{u}}}\cdot \frac{{du}}{{3\cancel{{\sec \theta \,\text{tan}\theta }}}}\\&=\frac{1}{3}\int{{\frac{{du}}{u}=}}\frac{1}{3}\ln \left| {3\sec \theta -1} \right|+C\end{align}$ |
| $ \int{{4{{{\sec }}^{2}}\left( {4x} \right)}}\,dx$ | $ \begin{align}u=4x&\\du=4\,dx;\,\,\,dx&=\frac{{du}}{4}\end{align}$ $ \begin{align}\int{{4{{{\sec }}^{2}}\left( {4x} \right)}}\,dx&=\cancel{4}\int{{{{{\sec }}^{2}}}}\left( u \right)\frac{{du}}{{\cancel{4}}}\\&=\tan u+C=\tan \left( {4x} \right)+C\end{align}$ (ln not needed!) |
| $ \displaystyle \int\limits_{{\frac{\pi }{2}}}^{\pi }{{\frac{{1+\cos x}}{{x+\sin x}}}}\,dx$ | $ \begin{align}u&=x+\sin x\\du&=\,\left( {1+\cos x} \right)dx\\dx&=\,\frac{{du}}{{1+\cos x}}\end{align}$ $ \displaystyle \begin{align}\int\limits_{{\frac{\pi }{2}}}^{\pi }{{\frac{{1+\cos x}}{{x+\sin x}}}}\,dx&=\int\limits_{{x=\frac{\pi }{2}}}^{{x=\pi }}{{\frac{{\cancel{{1+\cos x}}}}{u}}}\,\frac{{du}}{{\cancel{{1+\cos x}}}}=\int\limits_{{x=1}}^{{x=\pi }}{{\frac{{du}}{u}}}\,\\&=\left[ {\ln \left| u \right|} \right]_{{x=\frac{\pi }{2}}}^{{x=\pi }}=\left[ {\ln \left| {x+\sin x} \right|} \right]_{{\frac{\pi }{2}}}^{\pi }=\ln \left| {\pi +\sin \left( \pi \right)} \right|-\ln \left| {\frac{\pi }{2}+\sin \left( {\frac{\pi }{2}} \right)} \right|\\&=\ln \left| {\pi +0} \right|-\ln \left| {\frac{\pi }{2}+1} \right|=\ln \left| {\frac{\pi }{{\frac{{\pi +2}}{2}}}} \right|=\ln \left| {\frac{{2\pi }}{{\pi +2}}} \right|\approx .2005\end{align}$ |
Integrals of $ \boldsymbol {{{e}^{u}}}$ and $ \boldsymbol {{{a}^{u}}}$
The integrals of the exponential functions $ {{e}^{u}}$ and $ {{a}^{u}}$ mainly involves U-Substitution. When we take the integral of a base other than $ e$, we can either convert the function to base $ e$ using the formula $ {{a}^{x}}={{e}^{{\left( {\ln a} \right)x}}}$ (since $ \displaystyle {{e}^{{\left( {\ln a} \right)x}}}={{\left( {{{e}^{{\left( {\ln a} \right)}}}} \right)}^{x}}={{a}^{x}}$), or remember the formula below:
Integrals of $ \boldsymbol {{{e}^{u}}}$ and $ \boldsymbol {{{a}^{u}}}$
$ \int{{{{e}^{u}}}}={{e}^{u}}+C$ $ \displaystyle \int{{{{a}^{u}}}}du=\left( {\frac{1}{{\ln a}}} \right){{a}^{u}}+C$
Here are some Exponential Integration problems. It’s a good idea differentiate back to make sure you got the right answer!
First, the $ \boldsymbol {{{e}^{u}}}$ integration problems. Note that if there are multiple instances of $ {{e}^{x}}$ in the problem, we usually include $ {{e}^{x}}$ in the $ u$ part of the problem; otherwise we don’t.
| $ \boldsymbol {{{e}^{u}}}$ Integration Problem | Solution |
| $ \displaystyle \int{{{{e}^{{3x}}}}}\,dx$ | $ \begin{align}u=3x&\\du=3\,dx;\,\,\,\,dx&=\frac{1}{3}du\end{align}$ $ \displaystyle \int{{{{e}^{{3x}}}}}\,dx=\frac{1}{3}\int{{{{e}^{u}}\,du}}=\frac{1}{3}{{e}^{u}}+C=\frac{{{{e}^{{3x}}}}}{3}+C$ |
| $ \displaystyle \int{{{{e}^{x}}}}\sqrt{{3-{{e}^{x}}}}\,dx$ | $ \begin{array}{l}u=3-{{e}^{x}}\\du=-{{e}^{x}}\,dx\end{array}$ $ \displaystyle \begin{align}\int{{{{e}^{x}}{{{\left( {3-{{e}^{x}}} \right)}}^{{\frac{1}{2}}}}}}\,dx&=-\int{{{{u}^{{\frac{1}{2}}}}\,du}}=-\frac{2}{3}{{u}^{{\frac{3}{2}}}}+C\\&=-\frac{2}{3}{{\left( {3-{{e}^{x}}} \right)}^{{\frac{3}{2}}}}+C=-\frac{2}{3}\sqrt{{{{{\left( {3-{{e}^{x}}} \right)}}^{3}}}}+C\end{align}$ |
| $ \displaystyle \int{{\frac{{{{e}^{x}}+{{e}^{{-x}}}}}{{{{e}^{{-x}}}-{{e}^{x}}}}\,dx}}$ | $ \begin{align}u&={{e}^{{-x}}}-{{e}^{x}}\\du&=\left( {-{{e}^{{-x}}}-{{e}^{x}}} \right)\,dx\\dx&=\frac{{du}}{{-\left( {{{e}^{{-x}}}+{{e}^{x}}} \right)}}\end{align}$ $ \require {cancel} \displaystyle \begin{align}\int{{\frac{{{{e}^{x}}+{{e}^{{-x}}}}}{{{{e}^{{-x}}}-{{e}^{x}}}}\,dx}}=\int{{\frac{{\cancel{{{{e}^{x}}+{{e}^{{-x}}}}}}}{u}\,\cdot \frac{{du}}{{-\cancel{{\left( {{{e}^{{-x}}}+{{e}^{x}}} \right)}}}}}}\\=-\int{{\frac{{du}}{u}}}=-\ln \left| {{{e}^{{-x}}}-{{e}^{x}}} \right|+C\end{align}$ (Note that by making $ u$ the numerator didn’t work) |
| $ \displaystyle \int{{{{e}^{{-x}}}\csc \left( {{{e}^{{-x}}}} \right)}}\,dx$ | $ \begin{array}{l}u={{e}^{{-x}}}\\du=-{{e}^{{-x}}}dx\end{array}$ $ \displaystyle \begin{align}\int{{{{e}^{{-x}}}\csc \left( {{{e}^{{-x}}}} \right)}}\,dx&=-\int{{\csc u\,du}}=-\left( {-\ln \left| {\csc u+\cot u} \right|} \right)+C\\&=\ln \left| {\csc \left( {{{e}^{{-x}}}} \right)+\cot \left( {{{e}^{{-x}}}} \right)} \right|+C\end{align}$ |
| $ \displaystyle \int\limits_{1}^{4}{{\frac{{{{e}^{{\frac{4}{x}}}}}}{{{{x}^{2}}}}\,d}}x$ | $ \begin{align}u&=\frac{4}{x}=4{{x}^{{-1}}}\\du&=-4{{x}^{{-2}}}dx\\dx&=-\frac{{{{x}^{2}}}}{4}du\end{align}$ $ \begin{align}\int\limits_{1}^{4}{{\frac{{{{e}^{{\frac{4}{x}}}}}}{{{{x}^{2}}}}\,d}}x&=\int\limits_{{x=1}}^{{x=4}}{{\frac{{{{e}^{u}}}}{{\cancel{{{{x}^{2}}}}}}\cdot -\frac{{\cancel{{{{x}^{2}}}}}}{4}\,d}}u=-\frac{1}{4}\int\limits_{{x=1}}^{{x=4}}{{{{e}^{u}}\,du=}}\left[ {-\frac{{{{e}^{u}}}}{4}} \right]_{{x=1}}^{{x=4}}\\&=\left[ {-\frac{{{{e}^{{\frac{4}{x}}}}}}{4}} \right]_{1}^{4}=-\frac{e}{4}-\left( {-\frac{{{{e}^{4}}}}{4}} \right)=\frac{{{{e}^{4}}-e}}{4}\approx 12.970\end{align}$ |
And now, the $ \boldsymbol {{{a}^{u}}}$ integration problems:
| $ \boldsymbol {{{a}^{u}}}$ Integration Problem | Solution |
| $ \displaystyle \int{{{{4}^{x}}}}\,dx$ | $ \displaystyle \int{{{{4}^{x}}}}\,dx=\frac{{{{4}^{x}}}}{{\ln 4}}+C$ Alternate Method: $ \displaystyle \begin{align}\int{{{{4}^{x}}}}\,dx&=\int{{{{{\cancel{e}}}^{{\cancel{{\ln }}\left( {{{4}^{x}}} \right)}}}}}\,dx=\int{{{{e}^{{x\ln 4}}}}}\,dx=\int{{{{e}^{{\left( {\ln 4} \right)x}}}}}\,dx\\&=\frac{1}{{\ln 4}}{{e}^{{x\ln 4}}}+C=\frac{{{{e}^{{\ln {{4}^{x}}}}}}}{{\ln 4}}+C=\frac{{{{4}^{x}}}}{{\ln 4}}+C\end{align}$ |
| $ \displaystyle \int{x}\left( {{{2}^{{-2{{x}^{2}}}}}} \right)\,dx$ | $ \begin{align}u&=-2{{x}^{2}}\\du&=-4x\,dx\\dx&=-\frac{{du}}{{4x}}\end{align}$ $ \displaystyle \require {cancel} \begin{align}\int{x}\left( {{{2}^{{-2{{x}^{2}}}}}} \right)\,dx&=\int{{\cancel{x}\left( {{{2}^{u}}} \right)\cdot \,\left( {-\frac{{du}}{{4\cancel{x}}}} \right)}}\\&=-\frac{1}{4}\int{{{{2}^{u}}du}}=-\frac{{{{2}^{u}}}}{{4\ln 2}}+C=-\frac{{{{2}^{{-2{{x}^{2}}}}}}}{{4\ln 2}}+C\end{align}$ |
| $ \displaystyle \int\limits_{0}^{2}{{\left( {{{3}^{x}}-{{2}^{x}}} \right)}}\,dx$ | $ \displaystyle \begin{align}\int\limits_{0}^{2}{{\left( {{{3}^{x}}-{{2}^{x}}} \right)}}\,dx&=\left[ {\frac{{{{3}^{x}}}}{{\ln 3}}-\frac{{{{2}^{x}}}}{{\ln 2}}} \right]_{0}^{2}=\left( {\frac{{{{3}^{2}}}}{{\ln 3}}-\frac{{{{2}^{2}}}}{{\ln 2}}} \right)-\left( {\frac{{{{3}^{0}}}}{{\ln 3}}-\frac{{{{2}^{0}}}}{{\ln 2}}} \right)\\&=\frac{8}{{\ln 3}}-\frac{3}{{\ln 2}}\approx 2.954\end{align}$ |
Understand these rules and practice, practice, practice!
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