Evaluate: `int(x+1)sqrt(2x^2+3)dx`

Explanation

Let `{I}=int(x+1) sqrt{2 x^{2}+3} {dx} `

`{x}+1=lambda frac{d}{d x}(2 x^{2}+3)+mu `

`Rightarrow x+1=lambda(4 x)+mu`

Equating coefficient

`4 lambda=1`

`Rightarrow lambda=frac{1}{4}` and `mu=1`

`therefore I=int^{4}[frac{1}{4}(4 x)+1] sqrt{2 x^{2}+3} {dx}`

`=frac{1}{4} int(4 x) sqrt{2 x^{2}+3} {dx}+int sqrt{2 x^{2}+3} {dx}`

...