Q.Choose the correct answer: $\int \frac{dx}{e^x+e^{-x}}$ is equal to (A) $\tan^{-1}(e^x)+C$ (B) $\tan^{-1}(e^{-x})+C$ (C) $\log(e^x-e^{-x})+C$ (D) $\log(e^x+e^{-x})+C$

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{e^{4x}+e^{2x}}\,dx=$ (A) $\dfrac{1}{2}e^{x}\left(\sqrt{e^{2x}+1}\right)+\dfrac{1}{2}\operatorname{Sinh}^{-1}(e^{x})+c$ (B) $\dfrac{1}{2}e^{x}\left(\sqrt{e^{2x}+1}\right)+\operatorname{Sinh}^{-1}(e^{x})+c$ (C) $\dfrac{1}{2}\left(\sqrt{e^{2x}+1}\right)+\dfrac{1}{2}\operatorname{Sinh}^{-1}(e^{x})+c$ (D) $\sqrt{e^{4x}+e^{2x}}+\sqrt{e^{2x}+1}+c$

   ›Reveal solutionSolution

   Factor out $e^{2x}$ from under the root, substitute $t=e^x$, and apply the standard $\int\sqrt{t^2+1}\,dt$ formula. Answer matches option (A).

   Concept and Intuition

   Recognizing $e^{4x}+e^{2x}=e^{2x}(e^{2x}+1)$ turns the integrand into $e^x\sqrt{e^{2x}+1}$, and since $\dfrac{d}{dx}e^x=e^x$, the substitution $t=e^x$ converts this directly into the classic $\int\sqrt{t^2+1}\,dt$ form.

   Step-by-Step Solution

   1. $\sqrt{e^{4x}+e^{2x}}=\sqrt{e^{2x}(e^{2x}+1)}=e^x\sqrt{e^{2x}+1}$ (since $e^x>0$).
   2. Let $t=e^x\Rightarrow dt=e^x\,dx$. Then $\displaystyle\int e^x\sqrt{e^{2x}+1}\,dx=\int\sqrt{t^2+1}\,dt$.
   3. Standard result: $\displaystyle\int\sqrt{t^2+1}\,dt=\dfrac{t}{2}\sqrt{t^2+1}+\dfrac12\sinh^{-1}(t)+c$. …

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2. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\tanh(\log x) =$ (A) $\frac{x+1}{x-1}$ (B) $\frac{x^2+1}{x^2-1}$ (C) $\frac{x^2-1}{x^2+1}$ (D) $2x$

   ›Reveal solutionSolution

   Direct substitution of $t=\log x$ into the exponential definition of $\tanh$ gives $\dfrac{x^2-1}{x^2+1}$.

   Concept and Intuition

   $\tanh t$ is defined purely in terms of $e^t$ and $e^{-t}$, so whenever the argument is itself a

   logarithm, substituting $e^{\log x}=x$ turns the hyperbolic function into an ordinary algebraic

   expression in $x$ — no hyperbolic identities needed beyond the definition.

   Step-by-Step Solution

   1. Recall $\tanh t = \dfrac{e^t-e^{-t}}{e^t+e^{-t}}$.
   2. Substitute $t=\log x$: $e^t = e^{\log x} = x$, and $e^{-t}=1/x$.
   3. So $\tanh(\log x) = \dfrac{x-\frac1x}{x+\frac1x}$. …

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3. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x+1)\sqrt{x^2+1}} =$ (A) $\dfrac{1}{\sqrt{2}}\mathrm{Sinh}^{-1}\left(\dfrac{1+x}{1-x}\right) + c$ (B) $\dfrac{1}{\sqrt{2}}\mathrm{Sinh}^{-1}\left(\dfrac{1-x}{1+x}\right) + c$ (C) $-\dfrac{1}{\sqrt{2}}\mathrm{Sinh}^{-1}\left(\dfrac{1-x}{1+x}\right) + c$ (D) $-\dfrac{1}{\sqrt{2}}\mathrm{Sinh}^{-1}\left(\dfrac{1+x}{1-x}\right) + c$

   ›Reveal solutionSolution

   This tests the $x+a=1/t$ substitution trick for $\int \dfrac{dx}{(x+a)\sqrt{x^2+b^2}}$ type integrals, reducing them to an inverse-hyperbolic form. The answer is (C).

   Concept and Intuition

   When the integrand has a linear factor $(x+1)$ multiplying a quadratic square root $\sqrt{x^2+1}$, a direct trig/hyperbolic substitution on $x$ doesn't simplify the linear factor. The classic trick is to substitute the reciprocal of the linear factor, $x+1=\dfrac{1}{t}$, which turns the linear factor into $1/t$ and reshapes the quadratic under the root into another quadratic in $t$ — and crucially clears denominators so the whole thing becomes a standard $\int \dfrac{dt}{\sqrt{\text{quadratic in }t}}$, solvable via $\mathrm{Sinh}^{-1}$.

   Step-by-Step Solution

   1. Let $x+1=\dfrac1t$, so $x=\dfrac1t-1$ and $dx=-\dfrac{1}{t^2}dt$.
   2. Compute $x^2+1=\left(\dfrac1t-1\right)^2+1=\dfrac{1-2t+2t^2}{t^2}$, so $\sqrt{x^2+1}=\dfrac{\sqrt{2t^2-2t+1}}{t}$ (taking $t>0$).
   3. Substitute into the integral:

   $\int\frac{-dt/t^2}{\frac1t\cdot\frac{\sqrt{2t^2-2t+1}}{t}}=\int\frac{-dt/t^2}{\sqrt{2t^2-2t+1}/t^2}=-\int\frac{dt}{\sqrt{2t^2-2t+1}}.$

   4. Complete the square: $2t^2-2t+1=2\left(t-\tfrac12\right)^2+\tfrac12$, so $-\int\frac{dt}{\sqrt{2\left(t-\frac12\right)^2+\frac12}}=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{\left(t-\frac12\right)^2+\frac14}}=-\frac{1}{\sqrt2}\mathrm{Sinh}^{-1}(2t-1)+c.$ …

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4. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $x \in (-\infty, -1)$, then $\displaystyle\int \dfrac{1}{\sqrt{x^2-1}}\,dx =$ (A) $\log\left(x + \sqrt{x^2-1}\right) + c$ (B) $-\cosh^{-1}(-x) + c$ (C) $\log\left(-x + \sqrt{x^2-1}\right) + c$ (D) $\text{sech}^{-1} x + c$

   ›Reveal solutionSolution

   For $x<-1$ the standard antiderivative $\log(x+\sqrt{x^2-1})$ is undefined (argument negative), so the correct form must use $-x$; this works out to $-\cosh^{-1}(-x)+c$, verified by direct differentiation.

   Concept and Intuition

   The familiar formula $\displaystyle\int\dfrac{dx}{\sqrt{x^2-1}}=\log(x+\sqrt{x^2-1})+c$ is derived assuming $x>1$ (so that $x+\sqrt{x^2-1}>0$ and the log is defined). For $x<-1$, $x$ is negative and $\sqrt{x^2-1}<|x|=-x$, so $x+\sqrt{x^2-1}<0$ — the log of a negative number is undefined, and the antiderivative must be re-derived with the correct sign for this branch.

   Step-by-Step Solution

   1. For $x\in(-\infty,-1)$, substitute $x=-t$, where $t>1$ (so $dx=-dt$, $x^2-1=t^2-1$).
   2. $\displaystyle\int\dfrac{dx}{\sqrt{x^2-1}} = \int\dfrac{-dt}{\sqrt{t^2-1}} = -\log\left(t+\sqrt{t^2-1}\right)+c$ (using the standard formula, valid since $t>1$).
   3. Substitute back $t=-x$: $= -\log\left(-x+\sqrt{x^2-1}\right)+c$.
   4. Recall the definition $\cosh^{-1}(y) = \log\left(y+\sqrt{y^2-1}\right)$ for $y\ge1$. With $y=-x$ (valid since $x<-1\Rightarrow -x>1$): $\cosh^{-1}(-x) = \log\left(-x+\sqrt{x^2-1}\right)$.
   5. So the antiderivative is $-\cosh^{-1}(-x)+c$. …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $x > 0$, then $\displaystyle\int \dfrac{1}{\sqrt{x^4 + 2x^3 + 2x^2}}\, dx =$ (A) $-\dfrac{1}{\sqrt{2}}\, \mathrm{Sinh}^{-1}\left(\dfrac{x+2}{x}\right) + c$ (B) $\dfrac{1}{\sqrt{2}}\, \mathrm{Sinh}^{-1}\left(\dfrac{x+2}{x}\right) + c$ (C) $-\dfrac{1}{\sqrt{2}}\, \mathrm{Cosh}^{-1}\left(\dfrac{x+2}{x}\right) + c$ (D) $\dfrac{1}{\sqrt{2}}\, \mathrm{Cosh}^{-1}\left(\dfrac{x+2}{x}\right) + c$

   ›Reveal solutionSolution

   This is a $\int \dfrac{dx}{x\sqrt{ax^2+bx+c}}$ type integral that is best handled by the reciprocal substitution $x=1/t$. It reduces to a standard $\mathrm{Sinh}^{-1}$ form. Answer: $-\dfrac{1}{\sqrt2}\mathrm{Sinh}^{-1}\left(\dfrac{x+2}{x}\right)+c$.

   Concept and Intuition

   When the integrand has the pattern $x\sqrt{\text{quadratic in }x}$ under a root (not a perfect square times $x^2$), the substitution $x=1/t$ (so $dx=-dt/t^2$) often removes the awkward linear factor $x$ outside the root and leaves a clean quadratic-under-root integral in $t$, which is a standard $\mathrm{Sinh}^{-1}$ or $\mathrm{Cosh}^{-1}$ form depending on the sign of the discriminant-related constant.

   Step-by-Step Solution

   1. Factor: $x^4+2x^3+2x^2=x^2(x^2+2x+2)=x^2\big((x+1)^2+1\big)$. Since $x>0$, $\sqrt{x^4+2x^3+2x^2}=x\sqrt{(x+1)^2+1}$.
   2. The integral becomes $I=\displaystyle\int\frac{dx}{x\sqrt{(x+1)^2+1}}=\int\frac{dx}{x\sqrt{x^2+2x+2}}$.
   3. Substitute $x=\dfrac1t$, so $dx=-\dfrac{dt}{t^2}$. Then $x^2+2x+2=\dfrac{1}{t^2}+\dfrac2t+2=\dfrac{2t^2+2t+1}{t^2}$, so $\sqrt{x^2+2x+2}=\dfrac{\sqrt{2t^2+2t+1}}{t}$ (for $t>0$).
   4. $x\sqrt{x^2+2x+2}=\dfrac1t\cdot\dfrac{\sqrt{2t^2+2t+1}}{t}=\dfrac{\sqrt{2t^2+2t+1}}{t^2}$.
   5. So $I=\displaystyle\int\frac{-dt/t^2}{\sqrt{2t^2+2t+1}/t^2}=-\int\frac{dt}{\sqrt{2t^2+2t+1}}=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{t^2+t+\frac12}}$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x^2\left(\sqrt{1 + x^2}\right)}\,dx =$ (A) $\dfrac{-\sqrt{x^2 + 1}}{x} + c$ (B) $\dfrac{\sqrt{x^2 + 1}}{x} + c$ (C) $\dfrac{-\sqrt{x^2 - 1}}{x} + c$ (D) $\dfrac{\sqrt{x^2 - 1}}{x} + c$

   ›Reveal solutionSolution

   This is a memorizable standard form; substituting $x=\tan\theta$ confirms it. Answer: $-\dfrac{\sqrt{x^2+1}}{x}+c$.

   Concept and Intuition

   Integrals of the form $\int \dfrac{dx}{x^2\sqrt{x^2+a^2}}$ are handled cleanly with the trig substitution $x=a\tan\theta$, which turns $\sqrt{x^2+a^2}$ into $a\sec\theta$ and the whole integrand into a simple $\cos\theta/\sin^2\theta$ form.

   Step-by-Step Solution

   1. Let $x=\tan\theta$, so $dx=\sec^2\theta\,d\theta$ and $\sqrt{x^2+1}=\sec\theta$.
   2. The integral becomes $\displaystyle\int \frac{\sec^2\theta\,d\theta}{\tan^2\theta\sec\theta} = \int \frac{\sec\theta}{\tan^2\theta}\,d\theta = \int \frac{\cos\theta}{\sin^2\theta}\,d\theta$.
   3. With $u=\sin\theta$, $du=\cos\theta\,d\theta$: $\displaystyle\int \frac{du}{u^2} = -\frac1u = -\frac{1}{\sin\theta}$. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int_0^x \dfrac{t^2}{\sqrt{a^2+t^2}}\,dt=$ (A) $\dfrac{x}{2}\sqrt{a^2+x^2}+\log\left|x+\sqrt{a^2+x^2}\right|$ (B) $\sqrt{a^2+x^2}-a^2\sinh^{-1}\dfrac{x}{a}$ (C) $\dfrac{x}{2}\sqrt{a^2+x^2}+\dfrac{a^2}{4}\log\left|x+\sqrt{a^2+x^2}\right|$ (D) $\dfrac{x}{2}\sqrt{a^2+x^2}-\dfrac{a^2}{2}\sinh^{-1}\dfrac{x}{a}$

   ›Reveal solutionSolution

   This is the standard $\int\frac{t^2}{\sqrt{a^2+t^2}}dt$ form; evaluating it from 0 to $x$ gives $\frac{x}{2}\sqrt{a^2+x^2}-\frac{a^2}{2}\sinh^{-1}\frac{x}{a}$.

   Concept and Intuition

   This integral is a classic one usually derived via $t=a\sinh\theta$ or by writing $t^2=(a^2+t^2)-a^2$ and splitting; either way it reduces to a term from $\sqrt{a^2+t^2}$ plus a $\sinh^{-1}$ (equivalently log) term.

   Step-by-Step Solution

   1. Write $t^2=(a^2+t^2)-a^2$, so $\dfrac{t^2}{\sqrt{a^2+t^2}}=\sqrt{a^2+t^2}-\dfrac{a^2}{\sqrt{a^2+t^2}}$.
   2. $\displaystyle\int\sqrt{a^2+t^2}\,dt=\frac{t}{2}\sqrt{a^2+t^2}+\frac{a^2}{2}\sinh^{-1}\frac{t}{a}$ (standard result), and $\displaystyle\int\frac{dt}{\sqrt{a^2+t^2}}=\sinh^{-1}\frac{t}{a}$.
   3. Subtracting $a^2$ times the second from the first: $\int\frac{t^2}{\sqrt{a^2+t^2}}dt=\frac{t}{2}\sqrt{a^2+t^2}+\frac{a^2}{2}\sinh^{-1}\frac{t}{a}-a^2\sinh^{-1}\frac{t}{a}=\frac{t}{2}\sqrt{a^2+t^2}-\frac{a^2}{2}\sinh^{-1}\frac{t}{a}$. …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $u = \log_e\left[\tan\left(\dfrac{\pi}{4} + \dfrac{\theta}{2}\right)\right]$, then $\sinh u =$ (A) $\cos\theta$ (B) $\sec\theta$ (C) $\tan\theta$ (D) $\sin\theta$

   ›Reveal solutionSolution

   This is the classical Gudermannian/Mercator-projection identity relating $u=\log\tan(\pi/4+\theta/2)$ to $\theta$. Answer: $\sinh u = \tan\theta$.

   Concept and Intuition

   The function $u=\log\left[\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)\right]$ is the inverse Gudermannian function, famous from the Mercator map projection. The key algebraic fact is that $\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)$ simplifies exactly to $\sec\theta+\tan\theta$, which makes $e^u$ and $e^{-u}$ into simple secant/tangent combinations.

   Step-by-Step Solution

   1. Use the tangent addition formula: $\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right) = \dfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)}$.
   2. Multiply numerator and denominator by $\cos(\theta/2)$ and use half-angle identities; this expression is a well-known identity equal to $\sec\theta + \tan\theta$. (Quick check: at $\theta=0$, both sides equal $1$.)
   3. So $e^u = \tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right) = \sec\theta+\tan\theta$.
   4. Since $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta) = \sec^2\theta-\tan^2\theta = 1$, we get $e^{-u} = \sec\theta-\tan\theta$. …

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9. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.$\log(\cosh 3+\sinh 3) + \log(\cosh 3-\sinh 3) =$ (A) $1$ (B) $2$ (C) $3$ (D) $0$

   ›Reveal solutionSolution

   This tests the hyperbolic identity $\cosh^2\theta-\sinh^2\theta=1$ combined with the log product rule; the answer is $0$.

   Concept and Intuition

   Just as $\cos^2\theta+\sin^2\theta=1$ for circular functions, hyperbolic functions satisfy $\cosh^2\theta-\sinh^2\theta=1$ for every real $\theta$. Recognising the sum of two logs as the log of a product turns this into a one-line identity check.

   Step-by-Step Solution

   1. $\log(\cosh3+\sinh3)+\log(\cosh3-\sinh3)=\log\big[(\cosh3+\sinh3)(\cosh3-\sinh3)\big]$ by the product rule for logarithms.
   2. The bracket is a difference of squares: $(\cosh3+\sinh3)(\cosh3-\sinh3)=\cosh^23-\sinh^23$. …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.If $\tanh x = \text{sech}\, y = \dfrac{3}{5}$ and $e^{x+y}$ is an integer, then $e^{x+y} =$ (A) $2$ (B) $8$ (C) $1$ (D) $6$

    ›Reveal solutionSolution

    This tests converting hyperbolic-function values into exponentials and picking the integer root. Answer: $e^{x+y}=6$.

    Concept and Intuition

    $\tanh x$ and $\operatorname{sech} y$ can each be turned into an equation for $e^x$ (or $e^y$) using $\tanh x=\dfrac{e^{2x}-1}{e^{2x}+1}$ and $\cosh y=\dfrac{e^y+e^{-y}}{2}$. Once each exponential is found (possibly with two roots), multiply and use the "must be an integer" condition to select the right root.

    Step-by-Step Solution

    1. $\tanh x=\dfrac{e^{2x}-1}{e^{2x}+1}=\dfrac35$. Cross-multiplying: $5(e^{2x}-1)=3(e^{2x}+1) \Rightarrow 2e^{2x}=8 \Rightarrow e^{2x}=4 \Rightarrow e^x=2$ (positive root, as $e^x>0$).
    2. $\operatorname{sech}y=3/5 \Rightarrow \cosh y=5/3$.
    3. $\cosh y=\dfrac{e^y+e^{-y}}{2}=\dfrac53 \Rightarrow e^y+e^{-y}=\dfrac{10}{3}$.
    4. Let $t=e^y$: $t+\dfrac1t=\dfrac{10}{3} \Rightarrow 3t^2-10t+3=0$. …

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11. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.If $\cosh(x-\log 3) = \sinh x$, then $x=$ (A) $\dfrac{1}{2}\log 3$ (B) $\dfrac{1}{2}\log 6$ (C) $\dfrac{1}{2}\log 5$ (D) $\log 3$

    ›Reveal solutionSolution

    Expanding $\cosh(x-\ln3)$ via the addition formula and solving the resulting exponential equation gives $x=\tfrac12\ln6$.

    Concept and Intuition

    Hyperbolic functions of $\log k$ have clean closed forms ($\cosh(\log k)=\tfrac12(k+1/k)$, $\sinh(\log k)=\tfrac12(k-1/k)$), which lets us expand $\cosh(x-\ln3)$ using the standard hyperbolic subtraction formula and reduce the whole equation to one in $e^x$.

    Step-by-Step Solution

    1. $\cosh(\ln3)=\tfrac12(3+\tfrac13)=\tfrac{5}{3}$, $\sinh(\ln3)=\tfrac12(3-\tfrac13)=\tfrac{4}{3}$.
    2. $\cosh(x-\ln3)=\cosh x\cosh(\ln3)-\sinh x\sinh(\ln3)=\tfrac53\cosh x-\tfrac43\sinh x$.
    3. Set equal to $\sinh x$: $\tfrac53\cosh x-\tfrac43\sinh x=\sinh x \Rightarrow \tfrac53\cosh x=\tfrac73\sinh x \Rightarrow 5\cosh x=7\sinh x$. …

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12. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.$\sinh^{-1}\left(-\dfrac{4}{3}\right) =$ (A) $\log\left[\dfrac{\sqrt{7}-2}{3}\right]$ (B) $-\log 3$ (C) $\dfrac{1}{2}\log\left(\dfrac{1}{7}\right)$ (D) $-\log 4$

    ›Reveal solutionSolution

    This tests the logarithmic formula for the inverse hyperbolic sine. Direct substitution gives $-\log 3$.

    Concept and Intuition

    The inverse hyperbolic sine has a closed-form logarithmic expression, $\sinh^{-1}(x)=\log(x+\sqrt{x^2+1})$, derived by solving $x=\sinh y=\dfrac{e^y-e^{-y}}{2}$ for $y$. Plugging in the given value directly evaluates the expression.

    Step-by-Step Solution

    1. Use the identity $\sinh^{-1}(x) = \log\left(x + \sqrt{x^2+1}\right)$.
    2. Substitute $x=-\dfrac{4}{3}$: $x^2 = \dfrac{16}{9}$, so $x^2+1 = \dfrac{25}{9}$, and $\sqrt{x^2+1}=\dfrac{5}{3}$.
    3. So $x+\sqrt{x^2+1} = -\dfrac{4}{3}+\dfrac{5}{3} = \dfrac{1}{3}$.
    4. Hence $\sinh^{-1}\left(-\dfrac{4}{3}\right) = \log\left(\dfrac{1}{3}\right) = -\log 3$.

    Common Mistakes …

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