How do you integrate #int x^3sqrt(1+x^2)# using integration by parts?

Question

Christopher Allers · Accepted Answer

#((1+x^2)^(3/2)(3x^2-2))/15+C#

#I=intx^3sqrt(1+x^2)dx#

Before trying any integration by parts, we should attempt more basic substitutions to simplify the integral. Let #u=1+x^2#. This implies that #du=2xdx# and #x^2=u-1#.

Rearranging the integral:

#I=1/2intx^2sqrt(1+x^2)(2xdx)=1/2int(u-1)sqrtudu#

So, we see that integration by parts won't be necessary at all!

#I=1/2int(u^(3/2)-u^(1/2))du=1/2(u^(5/2)/(5/2)-u^(3/2)/(3/2))#

#color(white)I=1/2(2/5u^(5/2)-2/3u^(3/2))=u^(5/2)/5-u^(3/2)/3=(3u^(5/2)-5u^(3/2))/15#

#color(white)I=(u^(3/2)(3u-5))/15=((1+x^2)^(3/2)(3x^2-2))/15+C#

Christopher Allers · Answer

undefined To integrate $ \int x^3 \sqrt{1+x^2} $ using integration by parts, we choose $ u = x^3 $ and $ dv = \sqrt{1+x^2} \, dx $.

Then, $ du = 3x^2 \, dx $ and $ v = \int \sqrt{1+x^2} \, dx $.

To find $ v $, make the substitution $ x = \sinh(t) $, then $ dx = \cosh(t) \, dt $, and the integral becomes $ \int \cosh^2(t) \, dt $.

This integral can be solved using the identity $ \cosh^2(t) = \frac{1 + \cosh(2t)}{2} $.

After integrating and substituting back $ x $, we can then apply the integration by parts formula:

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

Substitute $ u $, $ v $, $ du $, and $ dv $ into the integration by parts formula and solve for the integral.