Question

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

Answer 1

looks like a tough one, give me 5 minutes

Answer 2

Is this the integral?

\[\int\limits \frac{ x-3 }{ (x ^{2} +2x+4)^2}\]

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

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]:


gets you home