Question

how to integrate arctan 3x.dx

Answer 1

http://answers.yahoo.com/question/index?qid=20071210222326AAxQbnL

Answer 2

thanks ;)

Answer 3

Here;s something else:
(another way)

Let 3x=tan(u) <- we need to make a triangle for this later
=>3 dx=sec^2(u) du

so we have
\[\int\limits_{}^{}\arctan(3x) dx=\int\limits_{}^{}\arctan(\tan(u)) \frac{1}{3} \sec^2(u) du\]
\[=\frac{1}{3}\int\limits_{}^{}u \sec^2(u) du\]
so now we do integration by parts!
\[=\frac{1}{3}(u \tan(u)-\int\limits_{}^{}\tan(u) du)+C\]
\[=\frac{1}{3} u \tan(u)-\frac{1}{3}\int\limits_{}^{}\frac{\sin(u)}{\cos(u)} du +C\]
we need a simple substitution here! :)
let b=cos(u) => db=-sin(u) du
\[=\frac{1}{3}u \tan(u)-\frac{1}{3}\int\limits_{}^{}\frac{-db}{b}+C\]
\[=\frac{1}{3} u \tan(u)+\frac{1}{3} \ln|b|+C\]
\[=\frac{1}{3}u \tan(u)+\frac{1}{3} \ln|\cos(u)|+C\]
now we need this back in terms of x
we call the first substitution
so we have 3x=tan(u)

We can find the hyp by using the Pythagorean thm
\[hyp=\sqrt{(3x)^2+1^2}=\sqrt{9x^2+1}\]
\[=\frac{1}{3} \tan^{-1}(3x) \cdot 3x+\frac{1}{3} \ln|\frac{1}{\sqrt{9x^2+1}}|+C\]
\[=x \tan^{-1}(3x)+\frac{1}{3} \ln|\frac{1}{\sqrt{9x^2+1}}|+C\]