Question

Integrate x/(x-1)(x^2+4) dx ?

Answer 1

`intx/((x-1)(x²+4))dx=1/5ln|x-1|-1/10ln(x²+4)+2/5tan^(-1)(x/2)+C`, `C in RR`

Explanation:

`intx/((x-1)(x²+4))dx`
Let `x/((x-1)(x²+4))=A/(x-1)+(Bx+C)/(x²+4)`
So:
`x=(A+B)x²+(C-B)x+4A-C`
By identification :
`A+B=0`[1]
`C-B=1`[2]
`4A-C=0`[3]

[3]`<-`[3]+[2]+[1]

`A+B=0`[1]
`C-B=1`[2]
`5A=1`[3]
`<=>`
`B=-1/5`
`C=4/5`
`A=1/5`

So:
`intx/((x-1)(x²+4))dx=int((1/5)/(x-1)+(-x+4)/(5(x²+4)))dx`
`=1/5int1/(x-1)dx-1/5int(x-4)/(x²+4)dx`
`=1/5ln|x-1|-1/10int(2x)/(x²+4)dx+4/5int1/(x²+4)dx`
`=1/5ln|x-1|-1/10ln(x²+4)+2/5tan^(-1)(x/2)+C`, `C in RR`
\0/ here's our answer !