Integral of:

$\displaystyle dx/(sqrt(x^2+2x+65)$

I completed the square on the denominator to rewrite it as:

$\displaystyle sqrt((x+1)^2+64)$

When I try to solve this, I let:

$\displaystyle u=x+1$

$\displaystyle du=dx$

So we get the integral of:

$\displaystyle du/sqrt(u^2+64)$

Then I set:

$\displaystyle u=8tan(theta)$

$\displaystyle du=8sec^2(theta)d(theta)$

This simplifies to just the integral of:

$\displaystyle sec(theta)d(theta)$

The integral of this is:

$\displaystyle ln(abs(tan(theta)sec(theta)) + C$

When I setup a right triangle, I get the following sides:

Hypotenuse:sqrt((x+1)^2+64)

Adjacenet: 8

Opposite: x+1

When I use these values to replace tan(theta) and sec(theta) my final answer is:

$\displaystyle ln(abs((x+1)/8) * (sqrt((x+1)^2+64)/8)) + C$

However, the online website has a different answer. It has the same basic terms but without dividing by 8. Where am I going wrong?