Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Any ideas how to solve it? $$\int\frac{x^4+2x+4}{x^4-1}dx$$ Thanks!

share|improve this question
    
By the way, the trick to notice that your integrand is $1+\frac {2x+5}{x^4-1}$ is to "add zero": $$\frac{x^4+2x+4}{x^4-1} = \frac{x^4 -1 + 1+2x+4}{x^4-1} = \frac{x^4 -1}{x^4 - 1} + \frac{1+2x+4}{x^4-1}. $$ –  JavaMan May 7 '13 at 15:45
add comment

3 Answers

up vote 4 down vote accepted

Using polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx = \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx $$

Expressing this as partial fractions, we need only find $A, B, C$

$$= \int \left(1 + \frac{A}{x+1} + \frac B{x-1} +\frac{C}{x^2 + 1}\right)\,dx$$

And then the integration is pretty straightforward.

share|improve this answer
    
I think is CX+D in the last fraction right? –  Ofir Attia May 7 '13 at 15:48
    
Ahhh...yes...fixed...! ;-) –  amWhy May 7 '13 at 15:49
    
$2x+5 = x^3(A+B+D)+x^2(B-A+C)+x(A+B-D)+1(B-A-C)$ this is the final equation , now need to find the parameters,thanks! –  Ofir Attia May 7 '13 at 15:57
    
Actually, you don't need $CX + D$ in the numerator here: it turns out that $C$ is sufficient. I used $$2x + 5 = A(x - 1)(x^2 +1) + B(x+1)(x^2 + 1) + C(x-1)(x+1)$$ and solved for $A$ when $x = -1$, for $B$ when $x = 1$, and for $C$ when $x = 0,$ given $A, B$ –  amWhy May 7 '13 at 16:13
    
Nice to get the feedback! +1 –  Amzoti May 8 '13 at 0:29
add comment

Hint $$ \frac{x^4+2x+4}{x^4-1} = 1 + \frac{2x+5}{x^4-1} = 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} $$ and use partial fractions.

share|improve this answer
add comment

Hint: First, make the integrand into $1+\frac {2x+5}{x^4-1}$ Now apply partial fractions.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.