回顾：指数积分法

(Videos) Integrating Exponential Functions

by PatrickJMT

Here is a video by PatrickJMT where he works through problems involving the integration of exponential functions.

The two integral formulas are: $$\boxed{\int e^x\,dx=e^x+C}\\ \boxed{\int a^x\,dx=\frac{a^x}{\ln a}+C}$$

Often, when integrating exponential functions, you use $u$-substitution, integration by parts, and/or a bit of algebra.

The first example he solves is: $$\int \frac{e^x+1}{e^x}\,dx$$ The trick here is to break up the fraction. $$\begin{align} &=\int \left[\frac{e^x}{e^x}+\frac{1}{e^x}\right]\,dx\\ &=\int \left[1+\frac{1}{e^x}\right]\,dx\\ &=\int \left[1+e^{-x}\right]\,dx\\ &=\int1\,dx+\int e^{-x}\,dx\\ \end{align}$$

To integrate $e^{-x}$ we can use a $u$-substitution. You might notice that this will just be $-e^{-x}$ but if not we will use $u$-substitution to show this.

Let $u=-x$. Then $du=-1\,dx$ or $dx=-1\,du$. Substituting these into the integral we get: $$\begin{align} &=\int1\,dx+\int e^{-x}\,dx\\ &=\int1\,dx+\int e^{u}(-1)\,du\\ &=\int1\,dx-\int e^{u}\,du\\ &=x-e^u+C\\ &=\boxed{x-e^{-x}+C} \end{align}$$

The second example is: $$\int 3^{\sin\theta}\cos\theta\,d\theta$$

In this example we start off right away with a $u$-substitution. Let $u=\sin\theta$ then $du=\cos\theta\,d\theta$. Plugging these in we get: $$\begin{align} &\int 3^{\sin\theta}\cos\theta\,d\theta\\ &=\int 3^{u}\,du\\ &=\frac{3^u}{\ln 3}+C\\ &=\boxed{\frac{3^{\sin\theta}}{\ln 3}+C} \end{align}$$

In this video, Patrick goes over two more examples.

The first example is: $$\int\frac{2^x}{2^x+1}\,dx$$

We start off with $u$-substitution. Let $u=2^x+1$. Then, $du=2^x\cdot\ln 2\,dx$ or $\frac{du}{\ln 2}=2^x\,dx$. We can substitute these in now. $$\begin{align} &\int\frac{2^x}{\color{red}{2^x+1}}\,dx\\ &=\int\frac{\color{red}{2^x}}{u}\,\color{red}{dx}\\ &=\int\frac{\frac{du}{\ln2}}{u}\\ &=\frac{1}{\ln2}\int\frac{1}{u}\,du\\ &=\frac1{\ln2}\ln|u|+C\\ &=\boxed{\frac1{\ln2}\ln|2^x+1|+C} \end{align}$$

The final example he looks at is: $$\int \frac{e^{\sqrt x}}{\sqrt x}\,dx$$

We can use $u$-substitution to solve this. Let $u=\sqrt x$. Then $du=\frac{1}{2\sqrt x}\,dx$ or $2\,du=\frac{1}{\sqrt x}\,dx$. We can substitute these in. $$\begin{align} &\int \frac{e^{\color{red}{\sqrt x}}}{\sqrt x}\,dx\\ &=\int \frac{e^u}{\color{red}{\sqrt x}}\,\color{red}{dx}\\ &=\int e^u(2)\,du\\ &=2\int e^u\,du\\ &=2e^u+C\\ &=\boxed{2e^{\sqrt x}+C} \end{align}$$