Solve the indefinite integral of ${\sqrt{x}\arctan\sqrt{x}\over 1+x}$

Question

$$\int {\sqrt{x}\arctan\sqrt{x}\over 1+x}dx$$

I tried a substituition of $x=t^2$, then doing it by parts. It didn't go too well...

Answer 1

with $t^2 = x,$ you have $$\int \frac{t\tan^{-1} t}{1+t^2} 2t\, dt = 2 \int \left(1 - \frac 1{1+t^2}\right)\tan^{-1}t \, dt=2\int\tan^{-1}t \, dt-\left(\tan^{-1}t\right)^2 $$

an integration by parts gives $$\int \tan^{-1}t\, dt= t \tan^{-1}t - \int\frac{t}{1+t^2}\, dt = t \tan^{-1}t-\frac 12\ln(1+t^2) + C$$

Answer 2

Set $\sqrt{x}=t$. Then $dx=2tdt$ and the integral becomes $\int \frac{2t^2\arctan t}{1+t^2}dt= \int \frac{2(1+t^2)-2}{1+t^2}\arctan t\, dt= \int 2\arctan t\,dt-2\int \frac{\arctan t}{1+t^2}dt= 2t \arctan t-\int \frac{2t}{1+t^2}-(\arctan t)^2= 2t \arctan t-\log(1+t^2)+(\arctan t)^2+c= 2\sqrt{x} \arctan\sqrt{x}-\log(1+x)+(\arctan\sqrt{x})^2+c$