`int (x + 1) sqrt(2x + x^2) dx` Evaluate the indefinite integral.

You need to use the following substitution  `2x + x^2= t` , such that:

`2x + x^2= t=>(2 + 2x)dx = dt => (x + 1) dx = (dt)/2`

`int (x+1)sqrt(2x+x^2)dx= (1/2)int sqrt t dt`

`(1/2)int sqrt t dt = (1/3)t^(3/2) + c`

Replacing back `2x + x^2` for t yields:

`int (x+1)sqrt(2x+x^2)dx = (1/3)(2x + x^2)^(3/2) + c`

Hence, evaluating the indefinite integral, yields `int (x+1)sqrt(2x+x^2)dx = (1/3)(2x + x^2)^(3/2) + c`