:::section[Recognizing Inverse-Hyperbolic Patterns] Many radical and rational integrals are derivatives of inverse hyperbolic functions. The parameter $a$ is assumed positive, and every formula must be read together with its real domain. ::: :::theorem[Standard Radical and Rational Forms] Let $a>0$. Then $$\begin{align*} \int \frac{dx}{\sqrt{x^2+a^2}} &=\sinh^{-1}\left(\frac{x}{a}\right)+C\\ &=\ln\left|x+\sqrt{x^2+a^2}\right|+C,\\ \int \frac{dx}{\sqrt{x^2-a^2}} &=\cosh^{-1}\left(\frac{x}{a}\right)+C, \qquad x>a,\\ \int \frac{dx}{a^2-x^2} &=\frac1a\tanh^{-1}\left(\frac{x}{a}\right)+C, \qquad |x|<a,\\ \int \frac{dx}{a^2-x^2} &=\frac1a\coth^{-1}\left(\frac{x}{a}\right)+C, \qquad |x|>a. \end{align*}$$ ::: :::proof[Proof of the First Formula] Let $$\begin{equation*} u=\frac{x}{a}. \end{equation*}$$ Then $dx=a\,du$ and $$\begin{align*} \int\frac{dx}{\sqrt{x^2+a^2}} &=\int\frac{a\,du}{a\sqrt{u^2+1}}\\ &=\int\frac{du}{\sqrt{u^2+1}}\\ &=\sinh^{-1}u+C. \end{align*}$$ Substituting back gives the result. The logarithmic form follows from the formula for $\sinh^{-1}$. $\square$ ::: :::key-formula[Forms with a Factor of x] For $a>0$, $$\begin{align*} \int \frac{dx}{x\sqrt{a^2+x^2}} &=-\frac1a\operatorname{csch}^{-1}\left|\frac{x}{a}\right|+C, \qquad x\ne0,\\ \int \frac{dx}{x\sqrt{a^2-x^2}} &=-\frac1a\operatorname{sech}^{-1}\left|\frac{x}{a}\right|+C, \qquad 0<|x|<a. \end{align*}$$ ::: :::solved-problem[An Inverse Hyperbolic Sine Integral] To determine $$\begin{equation*} \int\frac{dx}{\sqrt{x^2+16}}, \end{equation*}$$ compare with $a^2=16$, so $a=4$. Therefore $$\begin{equation*} \int\frac{dx}{\sqrt{x^2+16}} =\sinh^{-1}\left(\frac{x}{4}\right)+C. \end{equation*}$$ ::: :::solved-problem[A Rational Form] For $|x|<3$, $$\begin{align*} \int\frac{dx}{9-x^2} &=\frac13\tanh^{-1}\left(\frac{x}{3}\right)+C\\ &=\frac16\ln\left(\frac{3+x}{3-x}\right)+C. \end{align*}$$ ::: :::theorem[Integrals of Square Roots] Let $a>0$. Then $$\begin{align*} \int\sqrt{x^2+a^2}\,dx &=\frac{x}{2}\sqrt{x^2+a^2} +\frac{a^2}{2}\sinh^{-1}\left(\frac{x}{a}\right)+C,\\ \int\sqrt{x^2-a^2}\,dx &=\frac{x}{2}\sqrt{x^2-a^2} -\frac{a^2}{2}\cosh^{-1}\left(\frac{x}{a}\right)+C, \qquad x>a. \end{align*}$$ ::: :::mistake[One Algebraic Form Can Represent Different Branches] The integrand $1/(a^2-x^2)$ has different real inverse-hyperbolic antiderivatives on $|x|<a$ and $|x|>a$. A formula without its interval is incomplete. ::: :::exercise[Practice Questions] 1. Evaluate $\int dx/\sqrt{x^2+25}$. 2. Evaluate $\int dx/(16-x^2)$ for $|x|<4$. 3. Evaluate $\int dx/\sqrt{x^2-9}$ for $x>3$. 4. Differentiate the formula for $\int\sqrt{x^2+a^2}\,dx$. ::: :::answer[Answers and Guidance] 1. $\sinh^{-1}(x/5)+C$. 2. $\frac14\tanh^{-1}(x/4)+C$. 3. $\cosh^{-1}(x/3)+C$. 4. Apply the product rule and use $\frac{d}{dx}\sinh^{-1}(x/a)=1/\sqrt{x^2+a^2}$. ::: :::faq[Frequently Asked Questions] Q: Why must $a$ be positive? A: The standard forms use $a$ as a positive scale parameter and simplify $\sqrt{a^2}$ to $a$. Q: Why are absolute values needed in logarithmic forms? A: They keep the logarithm defined on each connected interval where the integrand is real. Q: When should $\tanh^{-1}$ be replaced by $\coth^{-1}$? A: Use $\tanh^{-1}$ for $|x|<a$ and $\coth^{-1}$ for $|x|>a$ in the integral of $1/(a^2-x^2)$. ::: :::call-to-action[Continue Learning] subtitle: Apply these formulas to powers, definite integrals, improper integrals, and mixed substitutions. button: Solve Hyperbolic Integration Applications|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/applications-of-hyperbolic-integration button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::