Question

Solve integral:

Answer 1

\[\int\limits x/(x^3-1)dx\]

Answer 2

you can replace x with sin x
then x= sinx
and \[x ^{2} -1 =\sin ^{2} -1 = \cos ^{2}\]
and dx=cos x dx

then the integration becomes
\[\int\limits_{}^{}sinx /\cos ^{2}x \times \cos x dx = \int\limits_{}^{}\tan x dx\]

Answer 3

i think its x^3

Answer 4

@El-naggar

Answer 5

yes x^3

Answer 6

i think better to use partial fractions....not sure....

Answer 7

sorry i saw it x^2

Answer 8

@El-naggar : it's okay

Answer 9

I got it. Gimmie 5 mins..

Answer 10

First split the equation into two, as thus:

\[{x \over (x-1)(x^2+x+1)} = {A \over (x-1)} + {Bx + C \over (x^2 + x + 1}\]
We get that
\[A(x^2 + x + 1) + (x-1)(Bx + C) = x\]
A = C, A= -B, B= -1/3
\[{x \over (x-1)(x^2+x+1)} = {1 \over 3}{1 \over (x-1)} + {1 \over 3}{-x + 1 \over (x^2 + x + 1)}\]
The integrals are now trivial, theyre both log!

Answer 11

oh i got it Thank again @soati :)

Answer 12

:) always a pleasure

Answer 13

:D