### A problem asked by Himanshu Sharma

Evaluate $\int \frac{2x+5}{(x+1)(x+2)(x+3)(x+4)+9} \; dx$

$(x+1)(x+2)(x+3)(x+4)+9$

$=(x^2 + 5x + 6)(x^2 + 5x + 4) + 9$

Substitute $u = x^2 + 5x + 5$ and $du = (2x + 5)\; dx$

And, we have

$\int \frac{2x+5}{(x+1)(x+2)(x+3)(x+4)+9} \; dx$

$=\int \frac{du}{(u+1)(u-1)+9}$

$=\int \frac{du}{u^2 + 8}$

$=\frac{1}{2\sqrt{2}}\tan^{-1} \frac{u}{2\sqrt{2}} + C$

$=\frac{1}{2\sqrt{2}}\tan^{-1} \frac{x^2 + 5x + 5}{2\sqrt{2}} + C$