Question

how do you find the integral of (6+9x)/(1+x^2)?

Answer 1

\[\large\rm \int\limits\frac{6+9x}{1+x^2}dx\]Hmm trig sub should work out nicely :)
\(\large\rm x=\tan\theta\)
ya?

Answer 2

I get this\[\int\limits_{}^{} \frac{ 6+9\tan \theta }{ 1+ \tan^2 \theta }\]

Answer 3

times sec^theta...so I guess the sec^theta cancel out

Answer 4

dtheta

Answer 5

\[\large\rm \int\limits \frac{6+9\tan \theta}{\cancel{1+\tan^2\theta}}\cancel{\sec^2\theta}~d \theta\]Ok good, that looks right so far.

Answer 6

How do you integrate tan theta?

Answer 7

Ummm, it ends up being some type of natural log.
You can convert to sines and cosines and apply a u-sub.\[\large\rm \int\limits 6~d \theta+\int \frac{\sin \theta}{\cos \theta}~d \theta\]
\(\large\rm u=\cos\theta\)