# Find the indefinite integral `int 5/(3e^x - 2) dx`

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### 2 Answers

The integral `int 5/(3*e^x - 2) dx` has to be determined.

Let `3*e^x - 2 = y`

`dy/dx = 3*e^x = y+2`

`int 5/(3*e^x - 2) dx`

= `int 5/(y*(y+2)) dy`

`1/(y*(y+2)) = A/y + B/(y+2)`

`1/(y*(y+2)) = (A*(y+2) + By)/(y*y+2))`

`Ay + 2A + By = 1`

2A = 1, A + B = 0

A = 1/2, B = -1/2

`int 5/(y*(y+2)) dy`

`= 5*int 1/(2*y) - 5/(2*(y+2)) dy`

`= (5/2)*ln y - (5/2)*ln(y+2) + C`

`= (5/2)*ln(3*e^x - 2) - (5/2)*ln(3*e^x) + C`

`= (5/2)*ln(3*e^x - 2) - (5/2)*ln 3 - (5/2)*ln e^x + C`

`= (5/2)*ln(3*e^x - 2) - (5/2)*x + C`

**The indefinite integral `int 5/(3*e^x - 2) dx = = (5/2)*ln(3*e^x - 2) - (5/2)*x + C` **

`int 5/(3e^x - 2) dx`

rewrite denominator as

`int 5/(e^x(3-2/e^x)) dx`

`int (5e^-x)/(3-2e^-x) dx`

Now we let `u = 3-2e^-x`

`du = 2e^-x dx`

`(du)/2 = e^-x dx`

Sub that in,

`int 5/(2u) du`

`5/2 int 1/u du`

`5/2 ln(u) + c`

`5/2 ln(3-2e^-x) + c`