Question

Who's bored enough to do a slightly tricky integral?\[\int\frac{x-3}{(4-2x-x^2)^2}dx\]

Answer 1

\[\int x^3+7x^2+17x+35 +\frac{121}{x-3} dx\]
\[= \frac{1}{4}x^4+\frac{7}{3}x^3+\frac{17}{2}x^2+25x +121\ln{(x-3)}+C\]

Answer 2

Here's the building blocks \[\int \frac{2t}{(1-t^2)^2} dt = \frac{1}{1-t^2} + C\]\[\int \frac{t^2+1}{(1-t^2)^2} dt = \frac{t}{1-t^2} + C \]\[\int \frac{t^2-1}{(1-t^2)^2} dt = \frac{1}{2}\log(1-t)+ \frac{1}{2}\log(1+t) \]

Answer 3

It should be 35x in the answer, typo :/

Answer 4

@Zed isn't that the integral of 1/F(x) not F(x)?

Answer 5

@Broken you have the first part of the integral going well, but how have you dealt with the -3 ?

Answer 6

I long divided the polynomials to make the integral easier

Answer 7

@Zed did you long divide upside-down o-0

Answer 8

ahhh I did, my apologies.

Answer 9

@TuringTest the constant term on top is obtained by taking the difference between integral 2 and integral 3, which has an integrand of 2/(stuff)^2

Answer 10

\[t=\frac{x+1}{\sqrt5}\] and compare the answer to WolframAlpha.

Answer 11

that's an interesting sub, I haven't checked wolf yet...

Answer 12

oops typo in integral 3 i think, 0.5*log((1-t)/(1+t)) gives the difference of two logs. otherwise it seems ok.

Answer 13

just about to say^
but yeah, it's the same otherwise, nice!

Answer 14

@Broken Fixer where did that sub come from?

Answer 15

its a straight up complete the square on the bottom, to convert it to 1-t^2 more amenable to trig substitution.

Answer 16

zed get your help

Answer 17

actually I guess I did the same thing for part of it, but I broke up the integral and used a trig sub on the integral with the constant on top
I used u=x+1 to start.
cool though, different ways are interesting ;-)

Answer 18

ahhh i see, interesting :)