Solution to $\int\frac{1}{\sqrt{1+4C+x^2}}$dx

Question

I would like to solve the following integral, but I am not sure if I am doing it the right way:

$$ \int \frac{1}{\sqrt{1+4C+x^2}}dx $$

with $C \in \mathbb{R}$.

I found this in an integral table: $$ \int\frac{dx}{\sqrt{a^2+x^2}}=\text{arcsinh}\left(\frac{x}{a}\right) $$
That suggests to identify $a^2$ with $1+4C$, such that $$ \int \frac{1}{\sqrt{1+4C+x^2}}dx = \text{arcsinh}\left(\frac{x}{\sqrt{1+4C}}\right) $$
Wolfram Alpha returns with $$ \int \frac{1}{\sqrt{1+4C+x^2}}dx = \text{arctanh}\left(\frac{x}{\sqrt{1+4C+x^2}}\right) $$
I am not really interested in that specific solution, but whether I was right in how I included $4C$ in the expression for $a^2$. Without the $4C$, the connection would have been straight forward: $a^2=1$ and $x^2=x^2$. Of course, the $4C$ has to go somewhere, and it cannot be the square term, so I included it in the constant term.

Am I allowed to do this?

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EDIT: If my solution with $\text{arcsinh}$ is correct, how can I prove equality with the one from W|A?

Answer 1

Everything depends on sign of $1+4C$. If $1+4C\gt 0$, then you can solve $a$ from $a^2=1+4C$ and have answer which found.

If $1+4C\lt 0$, then you have another integral. Use in this case $$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|x+\sqrt{x^2-a^2}|+B$$ Where $B$ is constant.

Last formula is, imho, more good, because generally is true $$\int\frac{dx}{\sqrt{x^2\pm a^2}}=\ln|x+\sqrt{x^2\pm a^2}|+B$$ for $a\gt 0$ and $B$ constant.

Addition for Edit:

Just differentiate right side.

Answer 2

Notice, if $1+4C\ge 0$ you can simply compare $\sqrt{1+4C+x^2}$ with $\sqrt{a^2+x^2}$ which gives you $1+4C=a^2$ or $a=\sqrt{1+4C}$

But for if $1+4C< 0$ it will be different.