How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.

$$\frac1{x+1+\sqrt{x^2+4x+5}}=-\frac{x+1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac{x+2-1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac12+\frac1{2(x+2)}+\frac{\sqrt{(x+2)^2+1}}{2(x+2)}$$
Setting $x+2=\tan y,$
$$\int\frac{\sqrt{(x+2)^2+1}}{(x+2)}\ dx=\int\frac{\sec y}{\tan y}\sec^2y\ dy$$
$$=\int\frac{dy}{\cos^2y\sin y}=\int\frac{\sin y\ dy}{\cos^2y(1-\cos^2y)}$$
Set $\displaystyle\cos y=u$
