Question

integrate x^3/(x-1) w.r.t x
How do i get the answer ?

Answer 1

first, just do a long division

x^3/(x-1) = x^2 + x + 1 + 1/(x-1)

Now just integrate each term.

Or, you could let u=x-1. Then you have

(u+1)^3/u = u^2 + 3u + 3 + 1/u
Integrate that and then return to x's. First way is easier, I think.

Answer 2

@steve ...i got the final ans as log y = 2[log(x-1)+(x-1)^3/3+3(x-1)+3/2(x-1)^2]
is it correct ?? did it by 2nd method

Answer 3

Did you check against the 1st method? That gives

x^3/3 + x^2/2 + x + log(x-1)


Your answer is clearly not the same, because right off you have 2log(x-1). Where did that 2 come from?

(x-1)^3/3+3(x-1)+3/2(x-1)^2
= x^3/3 + x^2/2 + x - 11/6

The 11/6 does not matter, since that's just part of C. But that 2log(x-1) is a killer.

Answer 4

@steve... ow... its ok... i did it in another sum where d 2 comes n mistakenly wrote it here... appreciate ur help :)