Another Integral

**Jgirl689** · February 24th 2010, 08:02 PM

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

**Prove It** · February 24th 2010, 08:08 PM

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.

**Locke** · February 24th 2010, 08:57 PM

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

**Prove It** · February 24th 2010, 10:17 PM

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.

**Jgirl689** · February 25th 2010, 07:27 PM

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

Thread: Another Integral