# Math Help - help plz

1. ## help plz

How would you solve this
integral of dx/(1+4x^2)
I know arctan is 1+x^2, but dont see how to solve this to make it look like arctan. thanks.

2. $4x^2 = (2x)^2$
So use the substitution $t=2x$.

3. 
\begin{aligned}
\int\frac1{1+4x^2}\,dx&=\frac14\int\frac1{\left(\d frac12\right)^2+x^2}\\&=\frac12\arctan2x+k,\,k\in\ mathbb R
\end{aligned}

Where $\color{red}\int\frac1{a^2+u^2}\,du=\frac1a\arctan\ frac ua+k,\,a\ne0$

4. Originally Posted by Krizalid

\begin{aligned}
\int\frac1{1+4x^2}\,dx&=\frac14\int\frac1{\left(\d frac12\right)^2+x^2}\\&=\frac12\arctan2x+k,\,k\in\ mathbb R
\end{aligned}

Where $\color{red}\int\frac1{a^2+u^2}\,du=\frac1a\arctan\ frac ua+k,\,a\ne0$
Krizalid, it should be $|a|$.