Question

How do you integrate `int5xsqrt(2x+3)` using substitution?

Answer 1

`int 5x * sqrt(2x+3) dx=1/2(2x+3)^(5/2) - 5/2(2x+3)^(3/2) + C`

Explanation:

This integral doesn't seem quiet obvious, so I will go through all the steps in high detail.

When evaluating integrals, it is sometimes a matter of trial and error, especially when it comes to u-substitutions.

For this particular integral, we may choose `u = 2x+3` and see what happens.

If `u=2x+3`, we'd have a `du` floating around somewhere - but there isn't. However, since there's a `5x` out front, we may be able to solve for `x`.

`u = 2x+3 -> du = 2 dx -> 1/2 du = dx`

Solving `u` for `x` we get

`u-3 = 2x -> (u-3)/(2) = x`

We can, however, pull out a `1/2` again, since it's just a constant.

Proceeding with the integration process, we can now rewrite our integral.

`int 5x * sqrt(2x+3) dx = 5 * 1/2 * 1/2 int (u-3) u^(1/2) du`

If we multiply everything into the parentheses and simplify we get

`5/4 int u^(3/2) - 3u^(1/2) du = 5/4[ 2/5u^(5/2) - 2/cancel(3) * cancel(3) u^(3/2)]+C`

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

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