What is ∫ e^x(1/x - 1/x^2) =?

2 Answers
Narad T.
Jun 10, 2018

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

Ayush S.
Jun 10, 2018

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

Related questions
Impact of this question
1363 views around the world
You can reuse this answer
Creative Commons License