Question

What is the integral of `int 1/(x^(3/2) + x^(1/2)) dx`?

Answer 1

`2arctan(x^(1/2))+C`

Explanation:

We have:

`I=int1/(x^(3/2)+x^(1/2))dx`

Factor the denominator.

`I=int1/(x^(1/2)(x+1))dx`

Which can be rewritten as:

`I=intx^(-1/2)/(x+1)dx`

`I=intx^(-1/2)/((x^(1/2))^2+1)dx`

Now, let `u=x^(1/2)`. Thus, `du=1/2x^(-1/2)dx`.

`I=2int(1/2x^(-1/2))/((x^(1/2))^2+1)dx`

Substituting:

`I=2int1/(u^2+1)du`

This is the arctangent integral:

`I=2arctan(u)+C`

`I=2arctan(x^(1/2))+C`