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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

kirkegaard
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    yes try partial fraction and show you work then someone will help you accordingly. – RE60K Mar 16 '15 at 10:57
  • KimR I am not one of the downvoters, but you should try to specify a bit on what your problem is. – AD - Stop Putin - Mar 16 '15 at 12:08
  • It's specified in the comments below. I can edit my question with my specific problem. Even though it's solved now. – kirkegaard Mar 16 '15 at 12:14
  • I prefer to guide the student, a solution does not help as much as finding the solution yourself. – AD - Stop Putin - Mar 16 '15 at 12:16
  • Very correct and i agree with you. The specifics i needed help with was the part where you get (Ax + B) + C. I'm used to only work with expressions where A + B + C. I'm not quite sure why i need to use the Ax + B yet. but when i did i solved the problem on my own. – kirkegaard Mar 16 '15 at 12:23

1 Answers1

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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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E.H.E
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  • Ok, i've done the partial fraction and got this: -5x + 11 = A(x^2+1) + B(x+1)

    And A = 8.. i struggle to find B..

     – kirkegaard Mar 16 '15 at 11:06
  • @KimR $\frac{-5x+11}{(x+1)(x^2+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{1+x}$ – E.H.E Mar 16 '15 at 11:08
  • Ok, i don't understand that part there. Can you please explain that? – kirkegaard Mar 16 '15 at 11:19
  • @KimR $$\frac{Ax+B}{x^2+1}+\frac{C}{1+x}=\frac{(Ax+B)(1+x)+C(x^2+1)}{(x^2+1)(1+x)}$$ – E.H.E Mar 16 '15 at 11:24
  • Ok, to be more specific; i dont understand how you get Ax + B. Sorry if i need this in with a tea-spoon but i feel that hard to grasp. Does it have something to do with the original expression (-5x +11)? – kirkegaard Mar 16 '15 at 11:35
  • @KimR, ok I will complete my answer – E.H.E Mar 16 '15 at 11:39
  • Ok, i've done some reading and found out when i should use Ax + B. (Heavesides Method) . Now i got the expression you wrote. That explained alot to me, thanks for your time @Hard – kirkegaard Mar 16 '15 at 12:13
  • The basic here is that $$\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd}+\frac{cb}{db} = \frac{ad+bc}{bd}$$ – AD - Stop Putin - Mar 16 '15 at 12:15