Answer
$\frac{-lnH}{H}-\frac{1}{H}=\frac{(1+R^{2})^{\frac{3}{2}}}{3}+C$
Work Step by Step
$\frac{dH}{dR}=\frac{RH^{2}\sqrt(1+R^{2})}{lnH}$
Rearrange the equation. Put all the H terms on the left side and all the R terms on the right.
$\frac{lnH}{H^{2}}dH=R\sqrt (1+R^{2})dR$
Now let's integrate both sides:
$\int \frac{lnH}{H^{2}}dH=\int R\sqrt (1+R^{2})dR$
This can be rewritten as:
$\int lnH(H^{-2})dH=\int R\sqrt (1+R^{2})dR$
On the left side, use integration by parts $\int u dv = uv-\int v du$ with $u=lnH$, $dv=\frac{1}{H^{2}}$, $du=\frac{1}{x}$, and $v=\frac{-1}{x}$.
On the right side, substitute using $u=1+R^{2}$ and $du=2x dx$.
We will have:
$\frac{-lnH}{H}-\int \frac{-1}{H^{2}}dH=\frac{1}{2}\int \sqrt u du$
Solve $\int \frac{-1}{H^{2}}dH$ and $\int \sqrt u du$ by applying the power rule. Then, we will have:
$\frac{-lnH}{H}-\frac{1}{H}=\frac{1}{2}(\frac{2u^{\frac{3}{2}}}{3})+C$
Undo substitution $u=1+R^{2}$:
$\frac{-lnH}{H}-\frac{1}{H}=\frac{(1+R^{2})^{\frac{3}{2}}}{3}+C$
