`int tln(t+1) dt` Find the indefinite integral

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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`

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