What is ∫ e^x(1/x - 1/x^2) =?

2 Answers
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#

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