# Thread: integral with dx on top?

1. ## integral with dx on top?

dx/(2 √x +2x)

I tried 1/(2 √x +2x)dx
then i know that it is not du/u so i tried finding u' but i don't think i have the right one...
i found 6x^1/2 because the bottom is 4x^3/2
so: -6x^1/2 6x^1/2/(2 √x +2x)dx
then: -6x^1/2[ln(abs value 4x^3/2)+c]

but i don't know after that.

the answer is ln(1+ √x) +c

2. Originally Posted by genlovesmusic09
dx/(2 √x +2x)

I tried 1/(2 √x +2x)dx
then i know that it is not du/u so i tried finding u' but i don't think i have the right one...
i found 6x^1/2 because the bottom is 4x^3/2
so: -6x^1/2 6x^1/2/(2 √x +2x)dx
then: -6x^1/2[ln(abs value 4x^3/2)+c]

but i don't know after that.

the answer is ln(1+ √x) +c
$\int\frac{\,dx}{2\sqrt{x}+2x}=\tfrac{1}{2}\int\fra c{\,dx}{\sqrt{x}+\left(\sqrt{x}\right)^2}$.

Let ${\color{red}u=\sqrt{x}}\implies\,du=\frac{\,dx}{2\ sqrt{x}}\implies 2\sqrt{x}\,du=\,dx$.

So we have $\tfrac{1}{2}\int\frac{2{\color{red}\sqrt{x}}\,du}{ u+u^2}=\int\frac{u\,du}{u\left(1+u\right)}=\int\fr ac{\,du}{1+u}$

Can you continue?