integrate x^3/(x-1) w.r.t x
How do i get the answer ?
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.
@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
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.
@steve... ow... its ok... i did it in another sum where d 2 comes n mistakenly wrote it here... appreciate ur help :)
To integrate the given expression, x^3/(x-1), with respect to x, we can follow these steps:
Step 1: Perform long division or divide x^3 by (x-1) to express it as a sum of a polynomial and a proper fraction:
x^3 / (x-1) = (x^2 + x + 1) + 1/(x-1)
Step 2: Integrate each term separately using the power rule for integration:
∫ (x^3 / (x-1)) dx = ∫ (x^2 + x + 1) dx + ∫ (1/(x-1)) dx
Step 3: Evaluate each integral term:
∫ (x^2 + x + 1) dx = (1/3)x^3 + (1/2)x^2 + x + C1, where C1 is the constant of integration.
∫ (1/(x-1)) dx = ln| x-1 | + C2, where C2 is the constant of integration.
Step 4: Combine the integration results:
∫ (x^3 / (x-1)) dx = (1/3)x^3 + (1/2)x^2 + x + ln| x-1 | + C
Therefore, the integral of x^3/(x-1) with respect to x is given as: (1/3)x^3 + (1/2)x^2 + x + ln| x-1 | + C, where C is the constant of integration.