Intergral Help!

**Nimmy** · January 16th 2006, 06:36 PM

Im struggling to find the integral of the following:

2
____________

(x+2)^2 (2-x)

Can someone show me the step by step solution. Plz thanks

**ThePerfectHacker** · January 16th 2006, 07:04 PM

Originally Posted by Nimmy

Im struggling to find the integral of the following:

2
____________

(x+2)^2 (2-x)

Can someone show me the step by step solution. Plz thanks

I presume you mean
$\int \frac{2dx}{(x+2)^2(2-x)}$
This type of problem requires fractional decomposition.
Thus, express
$\frac{2}{(x+2)^2(2-x)}=\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{C}{(2-x)}$
Now multiply the identity by $(x+2)^2(2-x)$
Thus,
$2=A(x+2)(2-x)+B(2-x)+C(x+2)^2$
Open sesame,
$2=x^2(-A+C)+x(4C-B)+(4A+2B+4C)$
Now that happens when
$-A+C=0$
$4C-B=0$
$4A+2B+4C=2$
Solving this we get that,
$(A,B,C)=(1/8,1/2,1/8)$
Thus,
$\frac{2}{(x+2)^2(2-x)}=\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}$
Thus,
$\int \frac{2dx}{(x+2)^2(2-x)}=\int (\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}) dx$
But,
$\int (\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}) dx$ = $\frac{1}{8}\ln |x+2|+\frac{1}{8}\ln |x-2|-\frac{1}{2}(x+2)^{-1}+C$
Q.E.D.

Note it is possible I made a mistake somewhere I am so tired right now and it took me a long time to write this in LaTeX form.

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