Question

Hi,
I would like to know the integration of
{(x-b) /((x^2 + 2bx - b^2)^2)).
Kindly help me to find this.

Answer 1

here u have to express (x-b) as A(diff of (x^2 + 2bx - b^2)) +B
where A and B are constants which we have to determine

Answer 2

this should be easy \[ \frac{x+ b }{((x + b)^2 - 2b^2)^2} -\frac{2 b }{((x + b)^2 - 2b^2)^2} \]
for the latter one, use trig subs

Answer 3

so (x-b) = A(2x +2b) +B
now equate the coefficients of x and the constant terms both for LHS and RHS
thus we have 2A=1 or A=1/2 and 2Ab +B =-b
or B = -b -2Ab
or B=-b-b=-2b

Answer 4

Hi experimentX,

Which trigonometric substitution could be effective here. I do not see any such type of substitution. Please explain.. Thanks

Answer 5

\[ \int \frac{x+b}{((x+b)^2 - 2b^2)^2 }dx = - \frac 1 2 \cdot \frac{1}{(x+b)^2 - 2b^2}\]
For the latter one, let (x+b)^2 = 2b^2sec^2(theta)
\[\int \frac{2 b }{((x + b)^2 - 2b^2)^2} dx = 2b\int \frac{ \sqrt 2 b \sec \theta \tan \theta }{4 b^2 \tan^4 \theta }d\theta \]

Answer 6

What about the first one..?

Answer 7

the first one, substitute, (x+b)^2 - 2b^2 = u

Answer 8

Ya, right..I will try and get back to u soon..Thanks a lot..:-)

Answer 9

you might have trouble with the last one
\[ \int \frac{\sec \theta \tan \theta}{\tan^4 \theta}d\theta = \int \frac{\sin^3 \theta}{\cos ^2 \theta} d\theta = \int \frac{(1 - \cos^2\theta) }{\cos^2 \theta} \sin \theta d\theta \]
after this, substitute, cos(theta) = u, you can do after this.

Answer 10

Ya,I understood.. Thanks a ton..