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This question already has an answer here:

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

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marked as duplicate by Did, user147263, ASB, Daniel W. Farlow, dustin Mar 21 '15 at 4:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ yes try partial fraction and show you work then someone will help you accordingly. $\endgroup$ – RE60K Mar 16 '15 at 10:57
  • $\begingroup$ KimR I am not one of the downvoters, but you should try to specify a bit on what your problem is. $\endgroup$ – AD. Mar 16 '15 at 12:08
  • $\begingroup$ It's specified in the comments below. I can edit my question with my specific problem. Even though it's solved now. $\endgroup$ – KimR Mar 16 '15 at 12:14
  • $\begingroup$ I prefer to guide the student, a solution does not help as much as finding the solution yourself. $\endgroup$ – AD. Mar 16 '15 at 12:16
  • $\begingroup$ 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. $\endgroup$ – KimR Mar 16 '15 at 12:23
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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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  • $\begingroup$ 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.. $\endgroup$ – KimR Mar 16 '15 at 11:06
  • $\begingroup$ @KimR $\frac{-5x+11}{(x+1)(x^2+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{1+x}$ $\endgroup$ – E.H.E Mar 16 '15 at 11:08
  • $\begingroup$ Ok, i don't understand that part there. Can you please explain that? $\endgroup$ – KimR Mar 16 '15 at 11:19
  • $\begingroup$ @KimR $$\frac{Ax+B}{x^2+1}+\frac{C}{1+x}=\frac{(Ax+B)(1+x)+C(x^2+1)}{(x^2+1)(1+x)}$$ $\endgroup$ – E.H.E Mar 16 '15 at 11:24
  • $\begingroup$ 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)? $\endgroup$ – KimR Mar 16 '15 at 11:35

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