integrate (t^2/(t+1)) dt?

Question

explain the procedure please

Answer 1

let's use a u substitution:
t + 1 = u, t = u - 1
dt = du

int((t^2)/(t+1)) = int((u - 1)^2/u)
int((u - 1)^2/u) = int (u^2 -2u + 1)/u
= int (u - 2 + 1/u)
(u^2)/2 - 2u + ln u + C

now resubstitute:
((t+1)^2)/2 - 2t - 2 + ln |t + 1| + C
(1/2)((t+1)^2 - 4t - 4) + ln |t + 1| + C
(1/2)(t^2 - 2t - 3) + ln |t + 1| + C
(1/2)(t - 3)(t + 1) + ln |t + 1| + C

Answer 2

Let u=t+1. Then the integral becomes ((u-1)^2/u)du (since du=dt and t=u-1). This expands to (u^2-2u+1)/u=u-2+1/u, which integrates to u^2/2-2u+log(abs(u))+C where log is the natural logarithm and abs is the absolute value, and C is any constant. Replacing u by t+1 gives (t+1)^2/2-2(t+1) + log(abs(t+1)) + C = (t^2+2t+1)/2-2t-2 + log(abs(t+1)) + C = t^2/2-t-3/2+log(abs(t+1)) +C = t^2/2-t + log(abs(t+1)) + C (I absorbed the constant -3/2 into the arbitrary constant C). This should be the answer.

Answer 3

t² / (t + 1) = t - t / (t + 1)
t² / (t + 1) = t - [1 - 1 / (t + 1) ]
t² / (t + 1) = t - 1 + 1 / (t + 1)
∫ t² / (t + 1) dt = t² / 2 - t + log (t + 1) + C

Answer 4

http://integrals.wolfram.com/index.jsp
try that link. its a function intergrater

Answer 5

no