Question

how to solve this integral (using partial fractions) ?

Answer 1

\[\int\limits \frac{ x }{ (x+1)(x^2-1) } ds\]

Answer 2

i took partial fractions as A/(x+1) + (Bx+C)/(x^2-1)

Answer 3

equating coefficient of x^2 : A+B=0
x : B+C =1
constant: C-A = 0

Answer 4

What I am thinking of, is (x+1)(x²-1)=(x+1)(x+1)(x-1)=(x+1)²(x-1)

And my partial fractions would be
-1/(4(x+1)
1/(2(x+1)²
1/4(x-1)

Answer 5

ok thanks for explaining... i will try it this way now

Answer 6

Sure...

Answer 7

@SolomonZelman to get A we do i have to compare the coefficient of x^2 terms ? but for B and C i can put values 1 and -1

Answer 8

I am not very sure what you mean

Answer 9

so the partial fraction would be A/(x+1) + B(x+1)^2 + C/((x-1)

Answer 10

think so.

Answer 11

Then you have a couple of steps left...

Answer 12

\[\large -\frac{1}{4}\int\limits_{ }^{ } \frac{1}{x+1}~dx + \frac{1}{2}\int\limits_{ }^{ } \frac{1}{(x+1)^2}~dx + \frac{1}{4}\int\limits_{ }^{ } \frac{1}{x- 1}~dx \]

Answer 13

You can sub some letters for the denominators or just apply integration without any subs... whatever you like

Answer 14

yes i solved it .. thank you very much

Answer 15

I am getting -log(x+1)/4 - 1/2(x+1) + log(x-1)

Answer 16

You can re-write `-log(x+1)/4 - 1/(2x+2) + log(x-1)` as a single fraction, but I don't think you have to, as far as I, (as a trig student) can imagine.

Answer 17

Sorry it should be

-log(x+1)/4 - 1/(2x+2) + log(x-1)/4

Answer 18

because the other one is also 1/4 § 1/(x-1) dx