Question

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

Answer 1

`(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`