Different methods of evaluating $\int\sqrt{a^2+x^2}dx$:

Question

I am looking for an elegant way like this Different methods of evaluating $\int\sqrt{a^2-x^2}dx$: to solve $\int\sqrt{a^2+x^2}dx$ without using trignometric identities such as x=atanp

Answer 1

You can pretty much recycle the derivation used in the answer you linked to. Start by integratring by parts, and rearrange:

$$\int \sqrt{x^2+1}\,dx = x\sqrt{x^2+1}-\int x\frac{x}{\sqrt{x^2+1}}dx\\ = x\sqrt{x^2+1}-\int\frac{x^2}{\sqrt{x^2+1}}dx\\ = x\sqrt{x^2+1}-\int\frac{x^2+1-1}{\sqrt{x^2+1}}dx\\ = x\sqrt{x^2+1}-\int\frac{x^2+1}{\sqrt{x^2+1}}dx + \int\frac{1}{\sqrt{x^2+1}}dx\\ = x\sqrt{x^2+1}-\int\sqrt{x^2+1}\,dx + \int\frac{1}{\sqrt{x^2+1}}dx\\ \implies 2\int \sqrt{x^2+1}\,dx = x\sqrt{x^2+1}+\int\frac{1}{\sqrt{x^2+1}}dx\\ \implies \int \sqrt{x^2+1}\,dx = \frac12x\sqrt{x^2+1}+\frac12\int\frac{1}{\sqrt{x^2+1}}dx\\ \implies \int \sqrt{x^2+1}\,dx = \frac12x\sqrt{x^2+1}+\frac12\sinh^{-1}{x}+C$$

Answer 2

Attempted different answer. Work with $x\ge 0, a \ge 0$. Write $$ F(a) = \int_0^u \sqrt{x^2+a^2}\;dx $$ for fixed $u$. Then compute $$ \frac{d}{dx} \;\sqrt{a^2+x^2} = \frac{a}{\sqrt{x^2+a^2}} $$ so $$ F'(a) = \int_0^u \frac{a}{\sqrt{x^2+a^2}}\;dx = -a\log a +a u\log\left(a + \sqrt{a^2+u^2}\right) $$ (assuming you know how to do this integral). Also $F(0) = u^2/2$. So integrate this with respect to $a$, use $F(0)$ to get the "constant": $$ F(a) = \frac{-\log \left( a \right) {a}^{2}+\log \left( u+\sqrt {{a}^{2}+{u}^{2}} \right) {a}^{2}+\sqrt {{a}^{2}+{u}^{2}}u}{2} $$ Check that $\frac{d}{du}$ of this is $\sqrt{a^2+u^2}$.