Integrals of the form ∫ xn ex dx by integration by parts (IBP)

These problems originally appeared in my former alternate blog at http://www.optimal-learning-systems.org/eps.


Q1. What is the value of the integral $\int xe^xdx$?

A1. Let $u=x$. Then $du=dx$. Also let $dv=e^xdx$. Then $v= \int e^xdx=e^x$.

Now apply the integration by parts formula, $\int u dv=uv−\int v du$.

We have, $\int x e^xdx=x e^x−\int e^xdx=x e^x−e^x=e^x[x−1]$.

Therefore, $\int xe^xdx=e^x[x−1]$.

One may wonder why the constant of integration is not added in the formula for v. The IBP formula becomes in this case $u(v+C)−∫(v+C)du$ and one can see that the terms containing the constant of integration cancels out.

Q2. Determine the integral $\int x^2e^x dx$.

A2. Let $u=x^2$, then $du=2^xdx$. Also let $dv=e^xdx$, then $v=\int e^xdx$. Then $dv=e^x dx$. Substituting in the IBP formula,

$x^2e^x−\int e^x(2x)dx=x^2e^x−2∫xe^xdx$. But the last term was the problem we treated before! Hence, $\int x^2e^xdx=x2e^x−2[e^x(x−1)]$ and therefore

$$\int x^2e^xdx=e^x [x2−2(x−1)]$$

You may add the constant of integration in the final answer.

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Q3. Determine the integral $\int x^3e^x dx$.

A3. As usual, let $u=x^3$, then $du=3x^2dx$. Also let $dv=e^xdx$, then $v=e^x$. Substituting in the IBP formula,

$x^3e^x−\int e^x(3x^2)dx=x^3e^x−3∫x2e^xdx$. But the last term involved an integral we treated before! Therefore $$\int x^3e^x dx=x^3e^x−3e^x[x^2−2(x−1)]=e^x[x^3−3(x^2−2(x−1))]$$
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Notice that $$\int x^ne^x dx=x^ne^x−n\int x^{n−1}e^xdx$$ . We leave the fun to the student to discover on his/her own a general formula for the answer.