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

Step-by-step solution

1. 1

   Recall the logarithmic form of arsinh. Solving $\sinh t=u$ for $t$ gives

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

   Note the $+1$ under the root: the standard form has no free parameter, so the given expression must first be scaled into that shape. This expression appears constantly as the antiderivative of $\dfrac{1}{\sqrt{a^2+x^2}}$, which is why the rewrite is worth knowing.

2. 2

   Scale the variable by $a$. Substituting $u=\dfrac{x}{a}$ into the identity:

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

3. 3

   Simplify the radical. Combining over $a^{2}$ and using $a>0$ so that $\sqrt{a^{2}}=a$:

   $\sqrt{\frac{x^{2}}{a^{2}}+1}=\sqrt{\frac{x^{2}+a^{2}}{a^{2}}}=\frac{\sqrt{x^{2}+a^{2}}}{a}$

   The assumption $a>0$ is essential here; for $a<0$ an absolute value would appear and the final constant would change.

4. 4

   Combine the two terms over $a$.

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

   using $\ln\frac MN=\ln M-\ln N$.

5. 5

   Rearrange to answer the question.

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

   The two expressions differ only by the additive constant $\ln a$, which is exactly why both appear as valid antiderivatives of $\dfrac{1}{\sqrt{a^{2}+x^{2}}}$ in different tables.

6. 6

   Verify numerically. With $a=2.7$ and $x=1.3$: the left side is $\ln(1.3+\sqrt{7.29+1.69})=1.4578391$, and $\operatorname{arsinh}(0.4814815)+\ln 2.7=0.4645873+0.9932518=1.4578391$ ✓.