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`
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