ज्ञात करें `int (x^2-x+1)/(""_((x^2 +1)^(3/2)))e^x dx`

Explanation

`int (x^2 -x+1)/(""_((x^2 +1)^(3/2)))e^x dx = int e^x [ (x^2+1)/(""_((x^2+1)^(3/2)))-x/(""_((x^2+1)^(3/2)))]dx`
`=int e^x [1/sqrt(x^2+1)+{(-x)/(""_((x^2+1)^(3/2)))}]dx`
`=int e^x[f(x)f'(x)]` जहां `f(x)=e^x/sqrt(x^2+1)+c`
`=e^x f(x) +c=e^x""1/sqrt(x^2+1)+c=e^x/sqrt(x^2+1)+c`