Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 15

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
For the given integral problem: int t ln(t+1) dt , we may apply u-substitution by letting:
u = t+1 that can be rearrange as t = u-1 .
The derivative of u is du= dt .
Plug-in the values, we get:
int t ln(t+1) dt= int (u-1) ln(u) du
Apply integration by parts: int f*g'=f*g - int g*f' .
We may let:
f =ln(u) then f' =(du)/u
g' =u-1 du then g=u^2/2 -u
Note: g =int g' = int (u+1) du .
int (u-1) du =int (u) du- int (1) du
= u^(1+1)/(1+1) - 1u
= u^2/2 - u
Applying the formula for integration by parts, we set it up as:
int (u-1) ln(u) du = ln(u) * (u^2/2-u) - int(u^2/2-u) *(du)/u
=(u^2ln(u))/2-u*ln(u) - int(u^2/(2u)-u/u) du
=(u^2ln(u))/2-u*ln(u) - int(u/2-1) du
For the integral part: int (u/2-1) du, we apply the basic integration property: int (u-v) dx = int (u) dx - int (v) dx .
int(u/2-1) du=int(u/2) du-int (1) du
= 1/2 int u - 1 int du
= 1/2*(u^2/2) - 1*u+C
= u^2/4 -u+C
Applying int(u/2-1) du=u^2/4 -u+C , we get:
int (u-1) ln(u) du =(u^2ln(u))/2-uln(u) - int(u/2-1) du
=(u^2ln(u))/2-u*ln(u) - [u^2/4 -u]+C
=(u^2ln(u))/2-u*ln(u) - u^2/4 +u+C
Plug-in u = t+1 on (u^2ln(u))/2-u*ln(u) - u^2/4 +u+C , we get the complete indefinite integral as:
int t ln(t+1) dt=((t+1)^2ln(t+1))/2-(t+1)ln(t+1) - (t+1)^2/4 +t+1+C
OR [(t+1)^2/2-t-1]ln(t+1) - (t+1)^2/4 +t+1+C