Evaluate indefinite integral. $\int\frac{dy}{\sqrt{169 + y^2}}$

Question

Evaluate $$\int\frac{dy}{\sqrt{169 + y^2}}$$

I have solved the problem, but don't seem to be getting the right answer. I get $\ln|\sqrt{13+y}+y|+C$ as the answer, and it is not the answer, how can I solve this?

Answer 1

There is another way to do this, one that I find highly interesting. Credit must go to @juantheron, whose post in another question first mentioned this.

Start by defining $\displaystyle x = \sqrt{y^2+a^2} \implies x^2 = y^2 + a^2 \implies x\,dx = y\,dy \implies \frac{dx}{y} = \frac{dy}{x}$

Using the simple algebra of proportions, we can see that $\displaystyle\frac{dx}{y} = \frac{dy}{x} = \frac{dx+dy}{x+y} = \frac{d(x+y)}{x+y}$.

Hence the original integral is $\displaystyle \int \frac{dy}{x} = \int \frac{d(x+y)}{x+y} = \ln |x+y| + C = \ln|y + \sqrt{y^2 + a^2}| + C$.

In your case, $\displaystyle a=13$.

Answer 2

Use the substitution $x=13\tan{u}$, then $dx=13\sec^2{u} \ du$ \begin{align} \int\frac{dx}{\sqrt{169+x^2}} &=\int\frac{13\sec^2{u} \ du}{\sqrt{169+169\tan^2{u}}}\\ &=\int\sec{u} \ du\\ &=\ln|\sec{u}+\tan{u}|+c\\ &=\ln\left|\frac{x}{13}+\sqrt{1+\left(\frac{x}{13}\right)^2}\right|+c \end{align} Alternatively, we can use the hyperbolic substitution $x=13\sinh{u}$, then $dx=13\cosh{u} \ du$. Thus \begin{align} \int\frac{dx}{\sqrt{169+x^2}} &=\int\frac{13\cosh{u} \ du}{\sqrt{169+169\sinh^2{u}}}\\ &=\int\frac{13\cosh{u} \ du}{13\cosh{u}}\\ &=\operatorname{arcsinh}\left(\frac{x}{13}\right)+c \end{align} These 2 answers are equal.