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.

i am trying to integrate following equation $$ \int\frac 1{(x^2-1)\cdot (x+2)}\,dx$$ i can represent $(x^2-1)=(x-1)(x+1)$ so,it would be converted in the following form $$\int\frac1{(x^2-1)(x+2)}\,dx=\int \frac1{(x-1)(x+1)(x+2)}\,dx$$ or it is equal $$\int \frac1{(x-1)(x^2+3x+2)}\,dx$$ last one we can decompose into form $$ \frac1{(x-1)(x^2+3x+2)}=\frac A{x-1}+\frac{Cx+D}{x^2+3x+2}$$ am i right?or did i miss some term?

share|improve this question
3  
Why don't you use $\frac 1{(x-1)(x+1)(x+2)}$ and decompose as $\frac \alpha{x-1} + \frac\beta{x+1} + \frac\gamma{x+2}$? –  martini Apr 22 '12 at 8:28
    
i was thinking that,because we have quadratic term,i would lose some fraction –  dato datuashvili Apr 22 '12 at 8:35
    
In partial fractions, you typically don't use quadratics in the denominator unless they can't be further reduced. –  Mike Apr 22 '12 at 8:53
    
thanks very much @Mike –  dato datuashvili Apr 22 '12 at 8:56
    
The common name for this type of integration is Integration using Partial Fractions. This is a special case of integrating fractions of the general form of $f(x)/g(x)$ –  Emmad Kareem Apr 22 '12 at 9:17

1 Answer 1

up vote 4 down vote accepted

I think you can decompose it like this:

$$ \frac{1}{(x^2-1)\cdot(x+2)}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{c}{x+2} $$

Thus we can solve the following equations:

$$ a+b+c=0\\3a+b=0\\2a-2b-c=1 $$

getting $a=1/6,b=-1/2,c=1/3$.

Therefore,

$$ \int\frac{dx}{(x^2-1)\cdot(x+2)}\\=\int\frac{1}{6}\cdot\frac{dx}{x-1}-\int\frac{1}{2}\cdot\frac{dx}{x+1}+\int\frac{1}{3}\cdot\frac{dx}{x+2}\\=\frac{1}{6}\cdot \log(x-1)-\frac{1}{2}\cdot \log(x+1)+\frac{1}{3}\cdot \log(x+2). $$

share|improve this answer
    
so it means that,my multiplication of (x+1) by (x+2) was not necessary,thanks a lot of @rhenskyyy –  dato datuashvili Apr 22 '12 at 8:51

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.