I have this problem $\int \frac {-5x + 11}{{(x+1)(x^2+1)}} \text{d}x$
And i'm not sure how to deal with this. I've tried substitution and getting nowhere. I've peeked at the answer and there a trigonometric part with arctan. Do i use Partial fraction expansion here?
Thanks
Answer
use the partial fraction to get
$$\frac{-5x+11}{(x+1)(x^2+1)}=\frac{3-8x}{x^2+1}+\frac{8}{x+1}=\frac{3}{x^2+1}-\frac{8x}{x^2+1}+\frac{8}{x+1}$$
details:
$$\frac{-5x+11}{(x+1)(x^2+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{1+x}=\frac{(Ax+B)(1+x)+C(x^2+1)}{(x^2+1)(1+x)}$$
$$\frac{-5x+11}{(x+1)(x^2+1)}=\frac{Ax+Ax^2+B+Bx+Cx^2+C}{(x^2+1)(1+x)}$$
$$A+B=-5$$
$$A+C=0$$
$$B+C=11$$
solve these simultaneous equations to get
$A=-8$
$B=3$
$C=8$
so the integral will be
$$\int (\frac{3}{x^2+1}-\frac{8x}{x^2+1}+\frac{8}{x+1})dx=3\tan^{-1}x-4\log(x^2+1)+8\log(x+1)+C $$
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