Solve Integral -$ \int \sqrt{3x^2 + 2x}\ dx$ I am trying to work out the integral without just skipping straight to the formula:
$$\int \sqrt{u^2-a^2} du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C   $$
I did all the substitutions, etc.
$$u=x+1/3,\,a=1/3,\,a\sec\theta=u,\,du=\sec\theta\tan\theta d\theta$$
and get
$$\sqrt3 \left [ \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|\frac{u}{a}+\frac{\sqrt{u^2-a^2}}{a}\right| \right] +C  $$
then
$$\sqrt3 \left [ \frac{1}{2}\left(x+\frac{1}{3}\right)\sqrt{x^2+\frac{2}{3}x} - \frac{1}{18} \ln\left|\left(3\left(x+\frac{1}{3}\right)+3\sqrt{x^2+\frac{2}{3}}\right)\right| \right] +C  $$
I have 2 questions -

*

*is my answer correct?

*And what happened to the $a$ under $\ln\left|\frac{u}{a}+\frac{\sqrt{u^2-a^2}}{a}\right|$ in the original general formula assuming $a\ne1$.

Thanks.

Yes your answer is correct. Here $a=\frac 13$, so $\frac 1a=3$, as you have written. Note that integration by parts can also be used to solve the problem.
