`int e^x/((e^(2x)+1)(e^x-1)) dx` Use substitution and partial fractions to find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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

Apply integral substitution:`u=e^x`

`=>du=e^xdx`

`=int1/((u^2+1)(u-1))du`

Now let's create partial fraction template for the integrand,

`1/((u^2+1)(u-1))=A/(u-1)+(Bu+C)/(u^2+1)`

Multiply the equation by the denominator,

`1=A(u^2+1)+(Bu+C)(u-1)`

`=>1=Au^2+A+Bu^2-Bu+Cu-C`

`=>1=(A+B)u^2+(-B+C)u+A-C`

Equating the coefficients of the like terms,

`A+B=0`    -------------------------(1)

`-B+C=0`  -----------------------(2)

`A-C=1`       -----------------------(3)

Now we have to solve the above three linear equations to get A, B and C,

From equation 1, `B=-A`

Substitute B in equation 2,

`-(-A)+C=0`

`=>A+C=0`    ---------------------(4)

Add equations 3 and 4,

`2A=1`

`=>A=1/2`

`B=-A=-1/2`

Plug in the value of A in equation 4,

`1/2+C=0`

`=>C=-1/2`

Plug in the values of A,B and C in the partial fraction template,

`1/((u^2+1)(u-1))=(1/2)/(u-1)+((-1/2)u+(-1/2))/(u^2+1)`

`=1/(2(u-1))-(1(u+1))/(2(u^2+1))`

`=1/2[1/(u-1)-(u+1)/(u^2+1)]`

`int1/((u^2+1)(u-1))du=int1/2[1/(u-1)-(u+1)/(u^2+1)]du`

Take the constant out,

`=1/2int(1/(u-1)-(u+1)/(u^2+1))du`

Apply the sum rule,

`=1/2[int1/(u-1)du-int(u+1)/(u^2+1)du]`

`=1/2[int1/(u-1)du-int(u/(u^2+1)+1/(u^2+1))du]`

Apply the sum rule for the second integral,

`=1/2[int1/(u-1)du-intu/(u^2+1)du-int1/(u^2+1)du]` ------------------(1)

Now let's evaluate each of the above three integrals separately,

`int1/(u-1)du`

Apply integral substitution:`v=u-1`

`dv=du`

`=int1/vdv`

Use the common integral:`int1/xdx=ln|x|`

`=ln|v|`

Substitute back `v=u-1`

`=ln|u-1|`    -------------------------------------------(2)

`intu/(u^2+1)du`

Apply integral substitution:`v=u^2+1`

`dv=2udu`

`int1/v(dv)/2`

Take the constant out and use standard integral:`int1/xdx=ln|x|`

`=1/2ln|v|`

Substitute back `v=u^2+1`

`=1/2ln|u^2+1|`    ----------------------------------------(3)

`int1/(u^2+1)du`

Use the common integral:`int1/(x^2+a^2)dx=1/aarctan(x/a)`

`=arctan(u)`  ------------------------------------------(4)

Put the evaluation(2 , 3 and 4) of all the three integrals in (1) ,

`=1/2[ln|u-1|-1/2ln|u^2+1|-arctan(u)]`

Substitute back `u=e^x` and add a constant C to the solution,

`=1/2[ln|e^x-1|-1/2ln|e^(2x)+1|-arctan(e^x)]+C`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial