How do you integrate #int x arctan x # using integration by parts?

Question

Emily Ammerman · Accepted Answer

#intxtan^{-1}(x)dx=1/2[(x^2+1)tan^{-1}(x)-x]+c#,
where #c# is the constant of integration.

#intxtan^{-1}(x)dx=1/2intfrac{d}{dx}(x^2)tan^{-1}(x)dx# #=1/2[x^2tan^{-1}(x)-intx^2frac{d}{dx}(tan^{-1}(x))dx]# #=1/2x^2tan^{-1}(x)-1/2intx^2(frac{1}{1+x^2})dx# #=1/2x^2tan^{-1}(x)-1/2int(1-frac{1}{1+x^2})dx# #=1/2x^2tan^{-1}(x)-1/2(x-tan^{-1}(x))+c#, where #c# is the constant of integration #=1/2[(x^2+1)tan^{-1}(x)-x]+c#

Samantha Adkison · Answer

undefined To integrate $ x \arctan(x) $ using integration by parts, you can use the formula:

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

Choose $ u = \arctan(x) $ and $ dv = x \, dx $, then differentiate $ u $ and integrate $ dv $ to find $ du $ and $ v $.

$$ du = \frac{1}{1 + x^2} \, dx $$
$$ v = \frac{1}{2}x^2 $$

Now apply the formula:

$$ \int x \arctan(x) \, dx = \frac{1}{2}x^2 \arctan(x) - \int \frac{1}{2}x^2 \frac{1}{1 + x^2} \, dx $$

The integral on the right side can be simplified by making a substitution $ u = 1 + x^2 $:

$$ \int \frac{1}{2}x^2 \frac{1}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C $$

$$ = \frac{1}{2} \ln(1 + x^2) + C $$

Therefore, the integral of $ x \arctan(x) $ using integration by parts is:

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