Question

Solve: $\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx$

Answer 1

$\begin{array}{c}\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx\\ =\int \left(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\right)\left(\frac{\sqrt{x}-\sqrt{1+x}}{\sqrt{x}-\sqrt{1+x}}\right)dx\\ =\int \left(\frac{{\left(\sqrt{1+x}\right)}^2+\sqrt{x}.\sqrt{1+x}}{{\left(\sqrt{1+x}\right)}^2-{\left(\sqrt{x}\right)}^2}\right)\left(\sqrt{1+x}-\sqrt{x}\right)dx\\ =\int \left(\frac{\left(\sqrt{1+x}\right)\left(\sqrt{1+x}+\sqrt{x}\right)}{1+x-x}\right)\left(\sqrt{1+x}-\sqrt{x}\right)dx\\ =\int \sqrt{1+x}(1+x-x)dx\\ =\int \sqrt{1+x}dx=\frac{2}{3}{(1+x)}^{\frac{2}{3}}+c\end{array}$