`int x ln(1 + x) dx` First make a substitution and then use integration by parts to evaluate the integral

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You need to use the substitution `1+x = t` , such that:

`1+x = t => dx = dt`

Changing the variable yields:

`int x*ln(1+x) dx = int (t-1)*ln t dt = int t*ln t dt - int ln t dt`

You need to use the integration by parts for `int t*ln t dt`   such that:

`int udv = uv - int vdu`

`u = ln t => du = 1/t`

`dv = t=> v = t^2/2`

`int t*ln t dt = ( t^2/2)*ln t - int (1/t)*(t^2/2) dt`

`int t*ln t dt = ( t^2/2)*ln t - (1/2) int t dt`

`int t*ln t dt = (t^2/2)*ln t - (t^2/4) + c`

You need to use the integration by parts for `int ln t dt`  such that:

`u =ln t=> du = 1/t`

`dv = 1=>v = t`

`int ln t dt = t*ln t - int t*1/t dt`

`int ln t dt = t*ln t - t + c`

`int (t-1)*ln t dt = (t^2/2)*ln t - (t^2/4) - t*ln t+ t + c`

Replacing back the variable, yields:

`int x*ln(1+x) dx = ((1+x)^2)/2*ln(1+x) - ((1+x)^2)/4 -(1+x)*ln (1+x) +(1+x) + c`

Hence, evaluating the integral, using  integration by parts, yields `int x*ln(1+x) dx = ((1+x)^2)/2*ln(1+x) - ((1+x)^2)/4 -(1+x)*ln (1+x) +(1+x) + c.`

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