`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...

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`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`