ज्ञात करें `int (xe^(x))/((1+x)^2)dx`

Explanation

`I = int (xe^(x))/((1+x)^2)dx = int (1+x-1)/((1+x)^2)e^x dx`
`=int e^x [(1+x)/((1+x)^2)-1/((1+x)^2)]ds = int e^x [1/(1+x)+{-1/((1+x)^2)}]dx`
`=int e^x [ f(x) +f'(x)] dx, ` जहां `f(x) = 1/(1+x)+c`
`=e^x f(x)+c=e^x""1/(1+x)+c=e^x/(1+x)+c`
`=e^x f(x) =+c=e^x""1/(1+x)+c=e^x/(1+x)+c`