How do you integrate #int x^3e^(x^2)# by integration by parts method?

Question

Christopher Allers · Accepted Answer

The integral is #(x^2-1)/2e^(x^2)+C#

First we use the substitution #u=x^2# so #du=2xdx# So the integral becomes #intx^3e^(x^2)dx# #=1/2intue^udu#

This is the integration by parts let #p=u# then #p'=1# and #v'=e^u# then #v=e^u#

#intpv'=pv-intp'v#

#1/2intue^udu=1/2(ue^u-inte^udu)# #=1/2(ue^u-e^u)#

#=(u-1)/2e^u#

Going back to x

#intx^3e^(x^2)dx=(x^2-1)/2e^(x^2) +C#

Charles Arocha · Answer

undefined To integrate $ \int x^3 e^{x^2} \, dx $ using integration by parts, we use the formula:

$$
\int u \, dv = uv - \int v \, du
$$

First, we choose:
- $ u = x^3 $, meaning $ du = 3x^2 \, dx $
- $ dv = e^{x^2} \, dx $, but to integrate $ dv $, we notice that the integral of $ e^{x^2} $ doesn't have a simple antiderivative in terms of elementary functions. However, recognizing that $ d(x^2) = 2x \, dx $, we can adjust our choice to work with this form. Let's adjust our strategy since the direct application of integration by parts with the initial choice doesn't lead to a straightforward solution.

Given the difficulty in directly applying integration by parts with the initial choice due to the nature of $ e^{x^2} $, a straightforward integration by parts isn't the most efficient method here. Instead, recognizing that the derivative of $ x^2 $ is $ 2x $ hints at a substitution method involving the exponential function that would be more appropriate.

To integrate $ \int x^3 e^{x^2} \, dx $ more effectively, we look for a substitution that aligns with the derivative of $ x^2 $, which is present in the exponential function. Here's a clearer path following the realization that our initial plan to directly apply integration by parts might not be the best approach:

Let $ u = x^2 $, then $ du = 2x dx $. Notice that our integral doesn't directly match this substitution because we have $ x^3 $ instead of $ x $. To adjust, we rewrite the integral in a form that matches our $ du $ substitution:

$$
\int x^3 e^{x^2} \, dx = \frac{1}{2} \int 2x \cdot x^2 e^{x^2} \, dx = \frac{1}{2} \int x^2 e^{u} \, (2x dx)
$$

With $ 2x dx = du $, we simplify to:

$$
\frac{1}{2} \int u e^{u} \, du
$$

Now, we can use integration by parts on this new form, where:
- Let $ u = u $ (to avoid confusion, let's refer to this as $ v $ moving forward) and $ dv = e^{u} \, du $
- Then, $ d(v) = du $ and $ v = e^{u} $

Applying integration by parts:

$$
\int v \, dv = v \cdot v - \int v \, d(v) = u e^{u} - \int e^{u} \, du = u e^{u} - e^{u} + C
$$

Substituting back $ u = x^2 $:

$$
\frac{1}{2} (x^2 e^{x^2} - e^{x^2}) + C
$$

Therefore, the integral is:

$$
\int x^3 e^{x^2} \, dx = \frac{1}{2} x^2 e^{x^2} - \frac{1}{2} e^{x^2} + C
$$