## Ac3 one year ago Evaluate the Integral of (x-3)/(x^2+2x+4)^2 please I'm stuck on this stupid problem.

1. alekos

looks like a tough one, give me 5 minutes

2. alekos

Is this the integral? $\int\limits \frac{ x-3 }{ (x ^{2} +2x+4)^2}$

3. Ac3

yes

4. Ac3

answer in back of book is (-1/2(x^2+2x+4))-(2sqrt(3)/9)arctan((x+1)/sqrt3)-(2(x+1))/(3(x^2+2x+4))+C

5. Ac3

I'm going to sleep I'll be back in about 8-9 hours.

6. IrishBoy123

recognising that $$\large \frac{d}{dx} (x^2 + 2x + 4) = 2(x+1)$$ you can start with : $$\large \int \frac{x-3}{(x^2 + 2x + 4)^2} \ dx= \int \frac{x+1}{(x^2 + 2x + 4)^2} - \frac{4}{(x^2 + 2x + 4)^2} \ dx$$ so the first part falls out simply as $$\large \frac{-1/2}{(x^2 + 2x + 4)}$$ for the second part, completing the square on the denominator leads to the rest: $$\large \int \frac{-4}{(x^2 + 2x + 4)^2} \ dx = \int \frac{-4}{((x + 1)^2 + \sqrt{3}^2)^2} \ dx$$ which suggests a substitution $$\large x + 1 = \sqrt{3} tan \theta, \ dx = \sqrt{3} sec^2 \theta \ d \theta$$ all of which simplifies eventually to: $$\frac{2 \sqrt{3}}{9} \int cos 2 \theta + 1 \ dx$$ integrating plus some double angle formula plus this [from the $$\theta$$ substitution]: |dw:1437734244042:dw| gets you home

7. alekos

good work @IrishBoy123. Heading in the right direction

8. alekos

by the way @Ac3 , this method leads to the answer that you've posted because I just completed the solution

9. Ac3

Thanks! Can't believe I didn't think of Trigonometric substitution.