How can you integrate #xe^(x^2) dx# using integration by parts?

Question

Charles Anctil · Accepted Answer

Although integrating by parts is not necessary, since it is specified,

We can integrate #xe^(x^2)dx# by sustitution #w=x^2# and we end up with

#int xe^(x^2)dx = 1/2 e^(x^2)+ C#

Therefore, to use parts, I will choose #u=1# and #dv=xe^(x^2)dx#

This makes #du=0dx# and #v= 1/2 e^(x^2)#

The components formula provides us with;

#1/2e^(x^2)- int 0 dx#

#= 1/2 e^(x^2) + C#

Charles Anctil · Answer

undefined To integrate $xe^{x^2} \, dx$ using integration by parts, you can choose $u = x$ and $dv = e^{x^2} \, dx$. Then, differentiate $u$ to get $du$ and integrate $dv$ to get $v$. Finally, apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$.

Adam Adkins · Answer

undefined To integrate $ xe^{x^2} $ using integration by parts, we select $ u = x $ and $ dv = e^{x^2} \, dx $. Then, we find $ du $ and $ v $ accordingly.

$$ u = x, \quad dv = e^{x^2} \, dx $$

$$ du = dx, \quad v = \int e^{x^2} \, dx $$

The integral $ \int e^{x^2} \, dx $ does not have a elementary antiderivative in terms of standard functions, so we cannot integrate it in this form. Therefore, we have to leave $ v $ as $ e^{x^2} $.

Applying the integration by parts formula:

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

$$ \int xe^{x^2} \, dx = \frac{1}{2}xe^{x^2} - \frac{1}{2} \int e^{x^2} \, dx $$

Now, we can rewrite the integral $ \int e^{x^2} \, dx $ in terms of $ xe^{x^2} $:

$$ \int xe^{x^2} \, dx = \frac{1}{2}xe^{x^2} - \frac{1}{2} \int e^{x^2} \, dx $$

$$ = \frac{1}{2}xe^{x^2} - \frac{1}{2} \cdot \frac{1}{2}e^{x^2} + C $$

$$ = \frac{1}{2}xe^{x^2} - \frac{1}{4}e^{x^2} + C $$

Where $ C $ is the constant of integration. Therefore, the integral of $ xe^{x^2} $ using integration by parts is $ \frac{1}{2}xe^{x^2} - \frac{1}{4}e^{x^2} + C $.