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) $\log(e^x+e^{-x})+C$

We are given:

and need to identify $f(x)$ from the options.

Concept and intuition:

The denominator $(\tan x + 1)^2$ and numerator containing $\sec^2 x + \tan x$ hint at a derivative of something like $\tan x$ or $x\tan x$. The given form of the answer suggests integration by parts was used, where one part became $\frac{-x^2}{x\tan x + 1}$ and the leftover integral is $f(x)$. Our job is to reconstruct that leftover integral by differentiating the given result.

Step-by-step reasoning:

1. Differentiate both sides Let $F(x) = \frac{-x^2}{x \tan x + 1} + f(x) + c$. Then $F'(x)$ must equal the integrand:

2. Differentiate the known part

   Let $g(x) = \frac{-x^2}{x \tan x + 1}$.

   Use quotient rule: $g(x) = \frac{-x^2}{u}$ where $u = x \tan x + 1$.

   Then $g'(x) = \frac{(-2x)u - (-x^2) u'}{u^2}$.

   Compute $u'$:

   $u = x \tan x + 1 \implies u' = \tan x + x \sec^2 x$.

   So:

3. Simplify the numerator of $g'(x)$ Expand:

So:

4. Relate to the integrand The integrand is:

Notice the denominator here is $(\tan x + 1)^2$, not $(x \tan x + 1)^2$. This is a mismatch — unless we rewrite the integrand cleverly.

Actually, the given answer’s denominator is $x \tan x + 1$, so the integration by parts likely used $x$ as one factor. Let’s instead directly differentiate the whole right-hand side and match.

5. Set up the equation Since $F'(x) = \text{integrand}$, we have:

So:

6. Find a common denominator

   Write $g'(x)$ with denominator $(\tan x + 1)^2$? But $g'(x)$ has denominator $(x \tan x + 1)^2$. This suggests the two denominators are different — unless we suspect a typo? Let’s check the given answer form: $\frac{-x^2}{x \tan x + 1}$ appears. That denominator is $x \tan x + 1$, not $\tan x + 1$. So the integration by parts likely used $u = x^2$, $dv = \frac{\sec^2 x + \tan x}{(\tan x + 1)^2} dx$.

   Let’s try that approach.

7. Integrate by parts

   Let $u = x^2$, $dv = \frac{\sec^2 x + \tan x}{(\tan x + 1)^2} dx$.

   Find $v$: Notice that $\frac{d}{dx}(\tan x + 1) = \sec^2 x$, so:

The first term is $-\frac{d}{dx}\left(\frac{1}{\tan x + 1}\right)$ because derivative of $(\tan x + 1)^{-1}$ is $-(\tan x + 1)^{-2} \sec^2 x$.

The second term: $\frac{\tan x}{(\tan x + 1)^2}$ can be written as $\frac{(\tan x + 1) - 1}{(\tan x + 1)^2} = \frac{1}{\tan x + 1} - \frac{1}{(\tan x + 1)^2}$.

So:

This is messy. Instead, note a simpler trick:

Observe that:

Not quite our numerator.

Better: Recognize that:

And also:

So the sum $\frac{\sec^2 x + \tan x}{(\tan x + 1)^2}$ is actually the derivative of something like $\frac{\tan x}{\tan x + 1} - \frac{1}{\tan x + 1} + \cdots$? Let’s check:

That gives $2\sec^2 x$, not our numerator.

Let’s try a direct approach:

Let $t = \tan x + 1$, then $dt = \sec^2 x \, dx$, but the $x^2$ outside makes substitution hard.

8. A better insight: Differentiate the given answer form

   Since the problem gives the result of integration, we can differentiate the RHS and compare. Let’s assume the given form is correct and find $f(x)$ by differentiating.

   Let $H(x) = \frac{-x^2}{x \tan x + 1} + f(x)$. Then $H'(x)$ equals the integrand.

   Compute derivative of $\frac{-x^2}{x \tan x + 1}$ as before:

Simplify numerator:

So:

Set equal to integrand:

9. Solve for $f'(x)$

This looks messy, but notice that the denominators are different. Perhaps the intended answer has $x \tan x + 1$ in denominator? Let’s check the options: they involve $\log|x\sin x + \cos x|$ or $\log|x\cos x + \sin x|$. Differentiate those:

• $\frac{d}{dx} \log|x\sin x + \cos x| = \frac{\sin x + x\cos x - \sin x}{x\sin x + \cos x} = \frac{x\cos x}{x\sin x + \cos x}$.
• $\frac{d}{dx} \log|x\cos x + \sin x| = \frac{\cos x - x\sin x + \cos x}{x\cos x + \sin x} = \frac{2\cos x - x\sin x}{x\cos x + \sin x}$.

None of these match a simple rational form with $(\tan x + 1)^2$. So maybe the integration by parts yields a different structure.

10. Try integration by parts directly Let $u = x^2$, $dv = \frac{\sec^2 x + \tan x}{(\tan x + 1)^2} dx$. Find $v$: Write:

Note that:

And:

So:

Integrate:

This is getting complicated. There must be a simpler pattern.

11. Spot the pattern: derivative of $\frac{x}{\tan x + 1}$

Compute:

Our numerator is $x^2(\sec^2 x + \tan x)$. If we factor $x$, we get $x \cdot x(\sec^2 x + \tan x)$. Not matching.

Try derivative of $\frac{x^2}{\tan x + 1}$:

That gives $2x \tan x + 2x - x^2 \sec^2 x$ over denominator. Our numerator is $x^2 \sec^2 x + x^2 \tan x$. So if we add something like $\frac{x^2 \tan x}{(\tan x + 1)^2}$? Not.

12. Consider the given answer’s denominator $x \tan x + 1$

That suggests the integration by parts used $u = x$, $dv = \frac{x(\sec^2 x + \tan x)}{(\tan x + 1)^2} dx$ perhaps. Let’s try:

Let $u = x$, $dv = \frac{x(\sec^2 x + \tan x)}{(\tan x + 1)^2} dx$.

Then $du = dx$, and we need $v$. Notice:

That’s not our $dv$. But:

Still not.

Try:

That’s close to our integrand’s numerator but with $x$ factor.

13. Direct differentiation of the given RHS

Let’s assume the given equation is correct and differentiate both sides to find $f(x)$ by matching.

Differentiate RHS:

Simplify numerator:

So:

Set equal to integrand:

Solve for $f'(x)$:

This seems too messy. Perhaps the problem has a misprint? But the options are simple logs, so $f'(x)$ must simplify to something like $\frac{2x \cos x}{x \sin x + \cos x}$ or similar.

14. Try to guess $f(x)$ from options

Differentiate option (C): $f(x) = 2\log|x\sin x + \cos x|$

Then:

Option (D): $f(x) = 2\log|x\cos x + \sin x|$  
Then:

Which one could appear from integration by parts? Notice that $\frac{d}{dx}(x\sin x + \cos x) = x\cos x$, so $f'(x)$ from (C) is $2 \cdot \frac{x\cos x}{x\sin x + \cos x}$. That is a neat derivative.

15. Check if the integrand can be expressed as derivative of $\frac{-x^2}{x\tan x + 1}$ plus that

Compute derivative of $\frac{-x^2}{x\tan x + 1}$ in a different form. Write $\tan x = \frac{\sin x}{\cos x}$, then:

So:

Differentiate this:

Numerator derivative: $(-2x\cos x + x^2 \sin x)(x\sin x + \cos x) - (-x^2 \cos x)(\sin x + x\cos x - \sin x)$ over denominator squared. …

For an even function $h$, $\displaystyle\int_{-k}^{k}\frac{h(x)}{1+a^{\,g(x)}}\,dx=\int_0^{k}h(x)\,dx$, because adding the substitution $x\to -x$ makes the two $a$-factors sum to $1$. Here $h(x)=\sec(2023x)$ is even and $g(x)=2023x$ with $a=2023$, over the symmetric interval $\left[-\frac{\pi}{8092},\frac{\pi}{8092}\right]$ (note $8092=4\cdot 2023$): …

The concept first: hunt for the $u$ whose derivative is already sitting there

Every successful substitution rests on the same observation: the integrand secretly contains $du$. So the productive question is not "what looks messy?" but "what could $u$ be, such that $du$ is also present?"

Here the options themselves are the clue — they are all built on $(1+\cos^{6}x)$ or $(1+\sec^{6}x)$ raised to a fractional power. That strongly suggests $u = 1+\cos^{6}x$, whose derivative is

So if we can massage the integrand into the shape $\sin x\cos^{5}x \times (\text{function of } u)$, we are done. That is exactly what converting $\sec \to \cos$ achieves.

Step-by-step

Step 1 — Express everything in cosines.

Step 2 — Rebuild the integrand. The denominator carries $(1+\sec^{6}x)$ to the power $2/3$:

Hence

Notice the payoff: exactly the $\sin x\cos^{5}x$ we predicted.

Step 3 — Substitute $t = \cos x$, so $dt = -\sin x\,dx$:

$I = -\int \frac{t^{5}}{\left(1+t^{6}\right)^{2/3}}\,dt$ …

Setup. We are given

The right-hand side is exactly the closed form of the inverse hyperbolic cosine:

Invert. Taking logarithms of the given relation,

hence

Differentiate.

Check by implicit differentiation. Differentiating $e^x = y+\sqrt{y^2-1}$: …

The problem asks us to find the equivalent expression for $\operatorname{Coth}^{-1} 2$ among the given options. The most straightforward way to do this is to convert $\operatorname{Coth}^{-1} 2$ into its logarithmic form and then convert each option into its logarithmic form to find a match. This approach works because inverse hyperbolic functions have well-defined logarithmic equivalents, which are often easier to compare.

Concept: Logarithmic Forms of Inverse Hyperbolic Functions

Inverse hyperbolic functions are defined in terms of the natural logarithm. This conversion is crucial for evaluating and comparing these functions. For example, if $y = \operatorname{Coth}^{-1} x$, it means $x = \operatorname{Coth} y$. By expressing $\operatorname{Coth} y$ in terms of exponential functions ($e^y$ and $e^{-y}$) and solving for $y$, we can derive its logarithmic form.

Let's proceed step-by-step to find the value of $\operatorname{Coth}^{-1} 2$ and then check the options.

Step-by-step Derivation

1. Calculate the value of $\operatorname{Coth}^{-1} 2$:

   We use the logarithmic form for $\operatorname{Coth}^{-1} x$.

   ›Proof

   Derivation of $\operatorname{Coth}^{-1} x$

   Let $y = \operatorname{Coth}^{-1} x$.

   Then, by definition, $x = \operatorname{Coth} y$.

   We know that $\operatorname{Coth} y = \frac{\cosh y}{\sinh y} = \frac{(e^y + e^{-y})/2}{(e^y - e^{-y})/2} = \frac{e^y + e^{-y}}{e^y - e^{-y}}$.

   So, we have the equation:

   $x = \frac{e^y + e^{-y}}{e^y - e^{-y}}$

   Multiply both sides by $(e^y - e^{-y})$:

   $x(e^y - e^{-y}) = e^y + e^{-y}$

   $xe^y - xe^{-y} = e^y + e^{-y}$

   Rearrange the terms to group $e^y$ and $e^{-y}$:

   $xe^y - e^y = xe^{-y} + e^{-y}$

   Factor out $e^y$ on the left and $e^{-y}$ on the right:

   $e^y(x - 1) = e^{-y}(x + 1)$

   Multiply both sides by $e^y$:

   $e^{2y}(x - 1) = (x + 1)$

   Isolate $e^{2y}$:

   $e^{2y} = \frac{x+1}{x-1}$

   Take the natural logarithm of both sides:

   $2y = \log\left(\frac{x+1}{x-1}\right)$

   Finally, solve for $y$:

   $y = \frac{1}{2} \log\left(\frac{x+1}{x-1}\right)$

   This formula is valid for $|x| > 1$.

   For $\operatorname{Coth}^{-1} 2$, we substitute $x=2$ into the formula:

   $\operatorname{Coth}^{-1} 2 = \frac{1}{2} \log\left(\frac{2+1}{2-1}\right)$

   $= \frac{1}{2} \log\left(\frac{3}{1}\right)$

   $= \frac{1}{2} \log 3$.

   Our target value is $\frac{1}{2} \log 3$. Now we check each option.

2. Evaluate Option (A): $\operatorname{Sech}^{-1} \left( \dfrac{1}{\sqrt{5}} \right)$:

   Using the formula $\operatorname{Sech}^{-1} x = \log\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ for $0 < x \le 1$. Here $x = \frac{1}{\sqrt{5}}$.

   $\operatorname{Sech}^{-1} \left( \dfrac{1}{\sqrt{5}} \right) = \log\left(\frac{1+\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^2}}{\frac{1}{\sqrt{5}}}\right)$

   $= \log\left(\frac{1+\sqrt{1-\frac{1}{5}}}{\frac{1}{\sqrt{5}}}\right)$

   $= \log\left(\frac{1+\sqrt{\frac{4}{5}}}{\frac{1}{\sqrt{5}}}\right)$

   $= \log\left(\frac{1+\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}\right)$

   $= \log\left(\frac{\frac{\sqrt{5}+2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}\right)$

   $= \log(\sqrt{5}+2)$.

   This does not match $\frac{1}{2} \log 3$. So, option (A) is incorrect. …