Rewrite ln(x + sqrt(a + x^2)) in terms of the inverse hyperbolic sine, assuming a > 0.

Step-by-step solution

1. 1

   Notice what is different about this version. The standard identity

   $\operatorname{arsinh}(u)=\ln\left(u+\sqrt{u^{2}+1}\right)$

   expects the constant under the root to be $1$. Here the constant is $a$ itself, not $a^{2}$, so the scale factor that normalises it is $\sqrt a$ rather than $a$ — and that single difference changes the additive constant at the end.

2. 2

   Factor $a$ out of the square root. For $a>0$,

   $\sqrt{x^{2}+a}=\sqrt{a\left(\frac{x^{2}}{a}+1\right)}=\sqrt a\,\sqrt{\left(\frac{x}{\sqrt a}\right)^{2}+1}$

   writing $\dfrac{x^{2}}{a}=\left(\dfrac{x}{\sqrt a}\right)^{2}$ to expose the required form.

3. 3

   Factor $\sqrt a$ out of the whole argument.

   $x+\sqrt{x^{2}+a}=\sqrt a\left(\frac{x}{\sqrt a}+\sqrt{\left(\frac{x}{\sqrt a}\right)^{2}+1}\right)$

   The bracket is now literally the argument of the arsinh identity with $u=\dfrac{x}{\sqrt a}$.

4. 4

   Split the logarithm of the product. Using $\ln(MN)=\ln M+\ln N$,

   $\ln\left(x+\sqrt{x^{2}+a}\right)=\ln\sqrt a+\operatorname{arsinh}\left(\frac{x}{\sqrt a}\right)$

5. 5

   Simplify the constant. Since $\ln\sqrt a=\ln a^{1/2}=\tfrac12\ln a$,

   $\ln\left(x+\sqrt{a+x^{2}}\right)=\operatorname{arsinh}\left(\frac{x}{\sqrt a}\right)+\frac12\ln a$

   Compare with the $a^{2}$ version, where the constant is a full $\ln a$: the exponent under the root is halved, and so is the log.

6. 6

   Verify numerically. With $a=2.7$, $x=1.3$: the left side is $\ln\left(1.3+\sqrt{2.7+1.69}\right)=1.2223723$, and $\operatorname{arsinh}\left(1.3/1.64317\right)+\tfrac12\ln 2.7=0.72575+0.49663=1.2223723$ ✓.