`inte^x/((e^x-1)(e^x+4))dx`

Let's apply integral substitution:`u=e^x`

`=>du=e^xdx`

`=int1/((u-1)(u+4))du`

Now create partial fraction template of the integrand,

`1/((u-1)(u+4))=A/(u-1)+B/(u+4)`

Multiply the above equation by the denominator,

`=>1=A(u+4)+B(u-1)`

`1=Au+4A+Bu-B`

`1=(A+B)u+4A-B`

Equating the coefficients of the like terms,

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

`4A-B=1` ----------------(2)

From equation 1, `A=-B`

Substitute A in equation 2,

`4(-B)-B=1`

`-5B=1`

`B=-1/5`

`A=-B=-(-1/5)`

`A=1/5`

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

`1/((u-1)(u+4))=(1/5)/(u-1)+(-1/5)/(u+4)`

`=1/(5(u-1))-1/(5(u+4))`

`int1/((u-1)(u+4))du=int(1/(5(u-1))-1/(5(u+4)))du`

Apply the sum rule,

`=int1/(5(u-1))du-int1/(5(u+4))du`

Take the constant out,

`=1/5int1/(u-1)du-1/5int1/(u+4)du`

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

`=1/5ln|u-1|-1/5ln|u+4|`

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

`=1/5ln|e^x-1|-1/5ln|e^x+4|+C`

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