Indifinite integration

Let $u = \sqrt{ax+b} \Rightarrow x = \frac{u^2 - b}{a} \Rightarrow dx = \frac{2u}{a} du$ . Now our original integral becomes:

$\int \frac{1}{ax + b + \sqrt{ax+b}} dx \Rightarrow \int \frac{1}{u^2 + u} \cdot \frac{2u}{a} du = \frac{2}{a} \int \frac{du}{u+1} = \frac{2}{a} ln(u+1) + C.$

A final re-substitutuion yields: $\boxed{\frac{2}{a} ln(\sqrt{ax+b} + 1) + C}.$

$\int\frac{1}{ax+b+\sqrt{ax+b}}dx$ $=\newline\int\frac{1}{\sqrt{ax+b}(1+\sqrt{ax+b})}dx$ $\newline$ Let $u=\sqrt{ax+b} \Rightarrow du=\frac{a}{2\sqrt{ax+b}}dx \Rightarrow dx= \frac{2}{a}udu$ $\newline$ So our integral becomes: $\newline\frac{2}{a}\int\frac{u}{u^{2}+u}du \newline =\frac{2}{a}\int\frac{1}{1+u}du \newline =\frac{2}{a}ln(u+1)+C \newline =\boxed{\frac{2}{a}ln(\sqrt{ax+b}+1)+C} (\because u=\sqrt{ax+b})$