# Math Help - Another Integral

1. ## Another Integral

Evaluate the indefinite integral. (Remember to use ln(abs(u)) where appropriate.)

2. Originally Posted by Jgirl689
Evaluate the indefinite integral. (Remember to use ln(abs(u)) where appropriate.)

$\int{\frac{e^{9x}}{e^{9x} + 3}\,dx} = \frac{1}{9}\int{\frac{9e^{9x}}{e^{9x} + 3}\,dx}$.

Now make the substitution $u = e^{9x}$ in the denominator.

3. u = e^9x
(1/9) du = e^9x dx
int((1 / u + 3)du)
ln|u + 3|
ln|e^9x + 3| + C

4. Originally Posted by Locke
u = e^9x
(1/9) du = e^9x dx
int((1 / u + 3)du)
ln|u + 3|
ln|e^9x + 3| + C
Correct.

But since $e^{9x} > 0$ for all $x$, you don't need the absolute value.

5. I put that in my answer thing, and it was marked wrong, should I use log instead?