The integral (1+x-1x)e^x+1x,dx is equal to

Question

The integral $\displaystyle\int \left(1+x-\dfrac1x\right)e^{x+\frac1x}\,dx$ is equal to

Accepted Answer

Let $I = \int \left(1+x-\dfrac1x\right)e^{x+1/x}\,dx = \int e^{x+1/x}\,dx + \int \left(x-\dfrac1x\right)e^{x+1/x}\,dx$

$= x\cdot e^{x+1/x} - \int x\left(1-\dfrac{1}{x^2}\right)e^{x+1/x}\,dx + \int\left(x-\dfrac1x\right)e^{x+1/x}\,dx$

$= xe^{x+1/x} - \int\left(x-\dfrac1x\right)e^{x+1/x}\,dx + \int\left(x-\dfrac1x\right)e^{x+1/x}\,dx = xe^{x+1/x}+C$

Solutions · 1