Question

Integrate:
$\int \frac{1}{\sqrt{x}+x}dx=$

Answer 1

$\begin{array}{cc}\int \frac{1}{\sqrt{x}+x}dx& \left[\begin{array}{c}put\sqrt{x}=u\\ diff.w.r.t.x,\\ \frac{1}{2\sqrt{x}}dx=du\\ \frac{1}{2u}dx=du\\ dx=2udu\end{array}\right]\\ \Rightarrow \int \frac{1}{u+u^2}.2udu& \\ \Rightarrow 2\int \frac{u}{u\left(u+1\right)}du& \\ \Rightarrow 2\int \frac{1}{u+1}du& \\ =2\log \left(u+1\right)\Rightarrow 2\log \left(\sqrt{x}+1\right)+c& \end{array}$