Integration help needed (tough one)

[dangerman77]

Can anyone get me started on this nasty integral:

(2x + 7)/(x^2 + 2x + 5)

-Completing the square doesn't work
-Can't divide it out

Mathematica gave me:
(5/2) ArcTan[(1 + x)/2] + Log[x^2 + 2x + 5)]

but I can't find how to get started.

 

[soroban]

Hello, dangerman77!

This one requires some gymnastics . . .

\(\displaystyle \L\:\int \frac{2x\,+\,7}{x^2\,+\,2x\,+\,5}\,dx\;=\;\int\frac{(2x\,+\,2)\,+\,5}{x^2\,+\,2x\,+\,5}\,dx\)

. . \(\displaystyle \L=\;\int\left(\frac{2x\,+\,2}{x^2\,+\,2x\,+\,5}\:+\:\frac{5}{x^2\,+\,2x\,+\,5}\right)\,dx\)

. . \(\displaystyle \L=\;\int\frac{2x\,+\,2}{x^2\,+\,2x\,+\,5}\,dx\:+\:5\int\frac{dx}{x^2\,+\,2x\,+\,5}\)


For the first integral, let \(\displaystyle u\,=\,x^2\,+\,2x\,+\,5\) . . . and we have: \(\displaystyle \L\int \frac{du}{u}\)


For the second integral, complete the square: \(\displaystyle \L\:5\int\frac{dx}{(x+1)^2\,+\,4}\)

. . and use: \(\displaystyle \L\;\int\frac{du}{u^2\,+\,a^2} \:=\:\frac{1}{a}\cdot\arctan\left(\frac{u}{a}\right)\,+\,C\)

 

[dangerman77]

Thank you! I had a feeling it would require some algebraic "gymnastics", and that first step completely eluded me. Cheers!