How do you integrate #(x^3)(x^2+1)^(1/2)#?

1 Answer
Jun 24, 2016

#(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+C#

Explanation:

We have:

#intx^3(x^2+1)^(1/2)dx#

We should try to use substitution. Let #u=x^2+1#, implying that #du=2xdx#.

Note that we don't visibly have #2xdx# in the integrand, but we can make some modifications.

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

Now, taking an #x# from the #x^3#, or splitting it into #x^3=x^2*x#:

#1/2intx^3(x^2+1)^(1/2)2dx=1/2intx^2(x^2+1)^(1/2)2xdx#

Here, we see that the leftover #x^2# can be expressed in terms of #u#: #x^2=u-1#.

#1/2intx^2(x^2+1)^(1/2)2xdx=1/2int(u-1)u^(1/2)du#

Now distributing the #u^(1/2)#:

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

Splitting into two integrals and integrating with the rule #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=-1#:

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

#=1/2(u^(5/2)/(5/2))-1/2(u^(3/2)/(3/2))+C=u^(5/2)/5-u^(3/2)/3+C#

#=(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+C#