Question

How to integrate the following:
2x/(x^3-1)

Answer 1

\[I=\int\limits \frac{2 x }{\left( x-1 \right)\left( x^2+x+1 \right) }dx\]
\[\frac{ 2x }{\left( x-1 \right)\left( x^2+x+1 \right) }=\frac{ A }{ x-1 }+\frac{ Bx+C }{ x^2+x+1 }\]
\[2x=A \left( x^2+x+1 \right)+\left( Bx+C \right)\left( x-1 \right)\]
equate terms like powers of x
0=A+B
2=A-B+C
0=A-C
find A,B,C and integrate.

Answer 2

can you complete it?

Answer 3

http://prntscr.com/8an7mo this term is much complicated

Answer 4

Thanks Surjithayer. Your reply was very helpful and I could complete it.

Answer 5

yw

Answer 6

Unable to get correct Answer. Can you further solve
To final answer.

Answer 7

https://www.symbolab.com/solver/definite-integral-calculator/%5Cint%5Cfrac%7B2x%7D%7B%5Cleft(x%5E%7B3%7D-1%5Cright)%7D%20dx/?origin=enterkey

Answer 8

A+B=0,B=-A
A-C=0
C=A
2=A+A+A
\[A=\frac{ 2 }{ 3 },B=-A=-\frac{ 2 }{ 3 },C=A=\frac{ 2 }{ 3 }\]

Answer 9

\[I=\int\limits \frac{ \frac{ 2 }{ 3 } }{ x-1 }dx+\int\limits \frac{ \frac{ -2 }{ 3 }x+\frac{ 2 }{ 3 } }{ x^2+x+1 }dx\]
\[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x-2 }{ x^2+x+1 }dx\]
\[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x+1-3 }{ x^2+x+1 }dx\]
\[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x+1 }{ x^2+x+1 }dx+\int\limits \frac{ 1 }{x^2+x+1 }dx\]

Answer 10

Thanks Surjithayer now it is clear.

Answer 11

yw