Question

What is ∫ e^x(1/x - 1/x^2) =?

Answer 1

The answer is `=e^x/x+C`

Explanation:

The integral is

`I=inte^x(1/x-1/x^2)dx`

`=inte^x/xdx-inte^x/x^2dx`

Perform the second integral by integration by parts

`intuv'=uv-intu'v`

`u=e^x`, `=>`, `u'=e^x`

`v'=1/x^2`, `=>`, `v=-1/x`

Therefore,

`I=inte^x/xdx-(-e^x/x+inte^x/xdx)`

`=inte^x/xdx+e^x/x-inte^x/xdx`

`=e^x/x+C`

Answer 2

`e^x1/x+C`
remember the answer as a useful property.

Explanation:

`inte^x(1/x-1/x^2)dx`

we have the property
`inte^x(f(x)+f'(x))dx=e^xf(x)+C`

let `f(x)=1/x` then, `f'(x)=-1/x^2`
therefore integral is of form `inte^x(f(x)+f'(x))dx`

so
`inte^x(1/x-1/x^2)dx=e^x1/x+C`

for proof of the property see below
`I=inte^x(f(x)+f'(x))dx=inte^xf(x)dx+inte^xf'(x)dx`
#=f(x) inte^xdx-inte^xf'(x)dx+ e^x intf'(x)dx-int e^xf(x)dx#
`=2e^xf(x)-inte^x(f(x)+f'(x))dx+K`
`=>I=2e^xf(x)-I+K`
`=>2I=2e^xf(x)+K`
`=>I=e^xf(x)+C`........,where`(C=K/2)`
remember this as a useful property