`int((x-1)e^x)/((x+1)^3)dx` का मान ज्ञात कीजिए ।

Explanation

`int((x-1)e^3)/((x+1)^3)dx=int(x-1)e^x(x+1)^(-3)dx`
`=(x-1)e^x int (x+1)^(-3)dx-int{d/(dx)(x-1)e^x int (x+1)^(-3)dx}dx`
`=(x-1)e^x[((x+1)^(-2))/(-2)]-int[d/(dx)(x-1)e^x.{((x+1)^(-2))/(-2)}]dx`
`=-1/2(x-1)e^x(x+1)^(-2)=1/2int[{(x-1)e^x=e^x1}(x+1)^(-2)]dx`
`=-1/2(x-1)(x+1)^2e^x=1/2intxe^x(x+1)^(-2)dx`
`=-1/2(x-1)(x+1)^(-2)e^x=1/2[xe^x.((x=1)^(-1))/(-1)-int(xe^x=e^x).((x+1)^(-1))/(-1)dx]`
`=-1/2(x-1)(x+1)^(-2)e^x-1/2xe^x(x+1)^(-1)+1/2int(e^x(x+1))/(x+1)dx`
`=-((x-1)e^x)/(2(x+1)^2)-(xe^x)/(2(x+1))+1/2inte^xdx`
`=-((x-1)e^x)/(2(x+1)^2)-(xe^x)/(2(x+1))=1/2inte^xdx`
`=-((x-1)e^x)/(2(x+1)^2)-(xe^x)/(2(x+1))=1/2e^x`