Looking for antiderivative

Question

Find $$\int\log\left(\sqrt{x+1}+\sqrt{x}\right)dx$$ I used the subtitution $$\sqrt{x}=\tan{y}$$ but i the integral became too long and involves $$\int\sec^3{y}dy$$ and also $$\int\sec^5{y}dy$$ but it solvable so i search for a simple solution if exist ? Note $$\log$$ is the natural logarithm

Answer 1

$$\int\ln(\sqrt{x+1}-\sqrt{x})dx=x\ln(\sqrt{x+1}-\sqrt{x})-\int \frac{x\left(\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}\right)}{\sqrt{x+1}-\sqrt{x}}dx=$$ $$=x\ln(\sqrt{x+1}-\sqrt{x})+\frac{1}{2}\int\sqrt{\frac{x}{x+1}}dx=...$$ and substitute $\sqrt{\frac{x}{x+1}}=t.$