Question

use substitution and partial fractions to find the indefinite integral. (remember to use absolute values where appropriate. use c for the constant of integration.) ex (ex − 17)(ex 10) d

Answer 1

I assume the integral is

`\displaystyle \int (e^x)/((e^x - 17) (e^x + 10)) \, dx`

Substitute
`y=e^x` and
`dy=e^x\,dx`.

`\displaystyle \int (dy)/((y - 17) (y + 10))`

Expand the integrand into partial fractions.

`\frac1{(y-17)(y+10)} = \frac a{y-17} + \frac b{y+10} \\\\ ~~~~ \implies 1 = a(y+10) + b(y-17) \\\\ ~~~~ \implies 1 = (a + b)y + 10a - 17b \\\\ ~~~~ \implies \begin{cases}a+b=0 \\ 10a-17b=1\end{cases} \\\\ ~~~~ \implies a=\frac1{27}, b=-\frac1{27}`

Then we have

`\displaystyle \frac1{27} \int \left(\frac1{y-17} - \frac1{y+10}\right) \, dy = \frac1{27} \left(\ln|y-17| - \ln|y+10|\right) + C \\\\ ~~~~~~~~ = \frac1{27} \ln\left|(y-17)/(y+10)\right| + C`

and putting the result back in terms of
`x`, we end up with

`\displaystyle \int (e^x)/((e^x - 17) (e^x + 10)) \, dx = \boxed{\frac1{27} \ln\left|(e^x - 17)/(e^x + 10)\right| + C}`