calculate:$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$

Question

calculate:$$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$$ I tried: $$\int \frac{x+1-\frac{\sqrt{x^2+1}}{x}+\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx=\int\frac{-dx}{x}+\int \frac{\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx$$ I can not continue from here

Answer 1

Let $x = \tan u $

$$\int \frac{(\tan u +1) \sec^2 u}{\sec u - \tan u ( \tan u +1 )} \; du $$ multiply with $\cos ^2 u $

$$\int \frac{(\tan u +1 )}{\cos u - \sin u ( \sin u + \cos u ) } $$

$$\int \frac{\sin u + \cos u }{\cos ^2 u - \sin u \cos u(\sin u + \cos u) } $$

$$\int \frac{2 \sqrt{2} \sin (0.25 \pi + u ) }{ 2\cos^2 u - \sin 2 u ( \sin ( 0.25 \pi +u))} \; du $$

$$\int \frac{ 2 \sqrt{2} ( \sin ( 0.25 \pi + u) ) }{\cos 2u +1 - \sin 2u ( \sin ( 0.25 \pi +u ))}$$ $\cos 2u = \cos 2 (y - 0.25 \pi) = \cos (2y - 0.5 \pi) = - \sin ( 2y) , \sin 2u = \sin(2y - 0.5 \pi ) = - \cos (2y)$ hence

$$\int \frac{2 \sqrt{2} \sin y}{- \sin 2y +1 + \cos 2y \sin y } $$

I hope this can help you. I will come back to check