Question

How do you evaluate `int(x)/(x^2+1)^(3/2) dx`?

Answer 1

`-1/sqrt(x^2+1)+C`

Explanation:

Through substitution, we can evaluate the indefinite integral. First, let

`u=x^2+1" "=>" "(du)/dx=2x" "=>" "du=2xdx`

Additionally, write `1/(x^2+1)^(3/2)` as `(x^2+1)^(-3/2)`.

This gives us the integral:

`intx(x^2+1)^(-3/2)dx=int(x^2+1)^(-3/2)*xdx`

Note that if we multiply the interior of the integral by `2`, we will have `2xdx`, our value for `du`. To do this, counterbalance it by multiply the exterior of the integral by `1/2`.

`=1/2int(x^2+1)^(-3/2)*2xdx`

Substituting in our values for `u` and `du`, this becomes:

`=1/2intu^(-3/2)color(white).du`

Now, evaluate the integral using the rule

`intu^ndu=u^(n+1)/(n+1)+C`

Giving the function:

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

`=1/2(-2)u^(-1/2)+C=-1/sqrtu+C`

Substituting in `u=x^2+1`, this yields

`=-1/sqrt(x^2+1)+C`