Answer
$$\int \frac{dx}{1+(3x+1)^2}=\frac{1}{3}\tan^{-1}(3x+1)+C$$
Work Step by Step
$$A=\int \frac{dx}{1+(3x+1)^2}$$
We set $a=3x+1$, which means $$da=3dx$$ $$dx=\frac{1}{3}da$$
Therefore, $$A=\frac{1}{3}\int\frac{da}{1+a^2}$$
We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$
So, $$A=\frac{1}{3}\times\frac{1}{1}\tan^{-1}\frac{a}{1}+C$$ $$A=\frac{1}{3}\tan^{-1}a+C$$ $$A=\frac{1}{3}\tan^{-1}(3x+1)+C$$
