How do I solve the integral $\int\frac{dz}{z\sqrt{4z+z^2}}$?

Question

The proposed integral is given by

$$\int \frac{dz}{z\sqrt{4z+z^2}}$$

I'm trying to answer this equation using the Reciprocal Substitution method, how should I approach this? Should I rewrite or simplify the equation first before I proceed?

If yes, someone please help me, an answer on this equation and as well an explanation would be so much appreciated too.

Thanks and advance and have a nice day.

Answer 1

Completing squares converts the integral to $$I=\int \frac{1}{z \sqrt{(z+2)^{2}-4}} d z.$$ Let $z+2=2 \sec \theta$, then $ d z=2 \sec \theta \tan \theta d \theta. $ $$ \begin{aligned} I &=\int \frac{2 \sec \theta \tan \theta d \theta}{2(\sec \theta-1) 2 \tan \theta} \\ &=\frac{1}{2} \int \frac{\sec \theta d \theta}{\sec \theta-1} \\ &=\frac{1}{2} \int \frac{1}{1-\cos \theta} d \theta \\ &=\frac{1}{2} \int \frac{1+\cos \theta}{\sin ^{2} \theta} d \theta \\ &=\frac{1}{2}\left[\int \csc ^{2} \theta d \theta+\int \csc \theta \cot \theta d \theta\right] \\ &=\frac{1}{2}(-\cot \theta-\csc \theta)+C \\ &=-\frac{1}{2}\left(\frac{z+4}{\sqrt{z^{2}+4 z}}\right)+C \\& \stackrel{OR}{=} -\frac{\sqrt{z^{2}+4 z}}{2 z}+C, \end{aligned} $$ which is checked by WA.

Answer 2

Hint: rewrite $$\int \frac{1}{z\sqrt{4z+z^2}}dz =\int \frac{1}{z^2\sqrt{\frac4z+1}}dz = -\int \frac{1}{\sqrt{\frac4z+1}}d(\frac1z)$$