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`int e^x/((e^x-1)(e^x+4)) dx` Use substitution and partial fractions to find the indefinite integral

`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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