Question

How do you integrate `int (2x^(3/2)+1)sqrtx` using substitution?

Answer 1

`int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C`

Explanation:

Our goal with substitution is to find an expression whose differential appears in the integral, and represent it using the variable `u.`

We have `int(2x^(3/2)+1)sqrtxdx`

In this case, `u=2x^(3/2)+1` is the best possible choice. Were we to choose `u=sqrtx, du=1/(2sqrtx)dx`, and this differential is not even close to anything in the integral.

Calculate its differential:

`du=(2)(3/2)x^(1/2)dx`

`du=2sqrtxdx`

Divide both sides by `2:`

`1/2du=sqrtxdx`

And we see `sqrtxdx` does in fact show up in the integral. So, rewrite with the substitution:

`1/2intudu=1/2(1/2)u^2+C=1/4u^2+C`

Rewriting in terms of `x` yields

`int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C`