Find: $\displaystyle\int \frac{e^{4x} - 1}{e^{4x} + 1}\, dx$

> [!TLDR] > The integral simplifies by rewriting the integrand as $\tanh(2x)$, whose antiderivative is $\frac{1}{2}\ln|\cosh(2x)| + C$. The final answer is $\boxed{\frac{1}{2}\ln(e^{4x}+1) - x + C}$. The key insight here is that the integrand $\frac{e^{4x} - 1}{e^{4x} + 1}$ is a disguised hyperbolic tangent. Recall that $\tanh u = \frac{e^{u} - e^{-u}}{e^{u} + e^{-u}} = \frac{e^{2u} - 1}{e^{2u} + 1}$. If we set $u = 2x$, then $e^{4x} = e^{2u}$, and the expression becomes $\frac{e^{2u} - 1}{e^{2u} + 1} = \tanh u = \tanh(2x)$. So the integral is $\int \tanh(2x)\, dx$. Why does this help? Because the derivative of $\cosh(2x)$ is $2\sinh(2x)$, and $\tanh(2x) = \frac{\sinh(2x)}{\cosh(2x)}$, so the integral is a natural logarithm of $\cosh(2x)$ up to a constant factor. This is the same trick as integrating $\tan x$: $\int \tan x\, dx = -\ln|\cos x| + C$. Let’s work through it step by step. 1. **Rewrite the integrand in hyperbolic form.** Multiply numerator and denominator of $\frac{e^{4x} - 1}{e^{4x} + 1}$ by $e^{-2x}$: $$ \frac{e^{4x} - 1}{e^{4x} + 1} = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \frac{\sinh(2x)}{\cosh(2x)} = \tanh(2x). $$ This is cleaner than working directly with exponentials. 2. **Set up the integral.** $$ I = \int \tanh(2x)\, dx = \int \frac{\sinh(2x)}{\cosh(2x)}\, dx. $$ 3. **Use substitution.** Let $u = \cosh(2x)$. Then $du = 2\sinh(2x)\, dx$, so $\sinh(2x)\, dx = \frac{du}{2}$. The integral becomes: $$ I = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C. $$ 4. **Back-substitute.** $$ I = \frac{1}{2} \ln|\cosh(2x)| + C. $$ Since $\cosh(2x) > 0$ for all real $x$, the absolute value is unnecessary: $$ I = \frac{1}{2} \ln(\cosh(2x)) + C. $$ 5. **Express in terms of exponentials (optional but often preferred in exam contexts).** Recall $\cosh(2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{e^{4x} + 1}{2e^{2x}}$. Then: …

Find: ∫ frace4x - 1e4x + 1 dx — CBSE Class 12 Mathematics 2024 PYQ

Question

\displaystyle\int \frac{e^{4x} - 1}{e^{4x} + 1}\, dx

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The integral simplifies by rewriting the integrand as $\tanh(2x)$ , whose antiderivative is $\frac{1}{2}\ln|\cosh(2x)| + C$ . The final answer is $\boxed{\frac{1}{2}\ln(e^{4x}+1) - x + C}$ .

The key insight here is that the integrand $\frac{e^{4x} - 1}{e^{4x} + 1}$ is a disguised hyperbolic tangent. Recall that $\tanh u = \frac{e^{u} - e^{-u}}{e^{u} + e^{-u}} = \frac{e^{2u} - 1}{e^{2u} + 1}$ . If we set $u = 2x$ , then $e^{4x} = e^{2u}$ , and the expression becomes $\frac{e^{2u} - 1}{e^{2u} + 1} = \tanh u = \tanh(2x)$ . So the integral is $\int \tanh(2x)\, dx$ .

Why does this help? Because the derivative of $\cosh(2x)$ is $2\sinh(2x)$ , and $\tanh(2x) = \frac{\sinh(2x)}{\cosh(2x)}$ , so the integral is a natural logarithm of $\cosh(2x)$ up to a constant factor. This is the same trick as integrating $\tan x$ : $\int \tan x\, dx = -\ln|\cos x| + C$ .

Let’s work through it step by step.

Rewrite the integrand in hyperbolic form. Multiply numerator and denominator of $\frac{e^{4x} - 1}{e^{4x} + 1}$ by $e^{-2x}$ :

\frac{e^{4x} - 1}{e^{4x} + 1} = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \frac{\sinh(2x)}{\cosh(2x)} = \tanh(2x).

This is cleaner than working directly with exponentials.

Set up the integral.

I = \int \tanh(2x)\, dx = \int \frac{\sinh(2x)}{\cosh(2x)}\, dx.

Use substitution. Let $u = \cosh(2x)$ . Then $du = 2\sinh(2x)\, dx$ , so $\sinh(2x)\, dx = \frac{du}{2}$ . The integral becomes:

I = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C.

Back-substitute.

I = \frac{1}{2} \ln|\cosh(2x)| + C.

Since $\cosh(2x) > 0$ for all real $x$ , the absolute value is unnecessary:

I = \frac{1}{2} \ln(\cosh(2x)) + C.

Express in terms of exponentials (optional but often preferred in exam contexts). Recall $\cosh(2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{e^{4x} + 1}{2e^{2x}}$ . Then: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$\int \frac{dx}{2^x + 2^{-x}}$
›Reveal solutionSolution

Rewrite the denominator as $2(2^{2x} + 1)/2^x$ , substitute $u = 2^x$ so $dx = \frac{du}{u \ln 2}$ , and recognize the arctangent integral form. The answer is $\frac{\tan^{-1}(2^x)}{\log 2} + C$ .

The key insight is to transform this exponential expression into a rational function that reveals an arctangent structure. The denominator $2^x + 2^{-x}$ looks symmetric, which suggests we can exploit the relationship between $2^x$ and $2^{-x}$ .

Start by rewriting the denominator in a more workable form. Multiply numerator and denominator by $2^x$ :
$\frac{1}{2^x + 2^{-x}} = \frac{2^x}{2^{2x} + 1}$
So our integral becomes:
$\int \frac{2^x \, dx}{2^{2x} + 1}$
Now the substitution becomes natural. Let $u = 2^x$ . Then:
$\frac{du}{dx} = 2^x \ln 2 = u \ln 2$
which gives us $dx = \frac{du}{u \ln 2}$ .

Substituting into the integral:
$\int \frac{u}{u^2 + 1} \cdot \frac{du}{u \ln 2} = \int \frac{1}{(u^2 + 1) \ln 2} \, du$
Factor out the constant:
$\frac{1}{\ln 2} \int \frac{du}{u^2 + 1}$
This is the standard arctangent integral. We know that $\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C$ .

Therefore: …
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CBSE 2025Set E1 markMCQ
$\int \frac{dx}{e^{-x}} =$
(a) $\frac{-1}{e^{-x}} + k$
(b) $e^x + k$
(c) $\frac{1}{e^{-x}} \cdot \frac{1}{x^2} + k$
(d) $-e^{-x} + k$
›Reveal solutionSolution

Rewrite $\frac{1}{e^{-x}}$ as $e^{x}$ and integrate; result $e^x+k$ .

Using $\frac{1}{e^{-x}}=e^{x}$ : …

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CBSE 2025Set E1 markMCQ
$\int e^3 \cdot e^x\,dx =$
(a) $e^x + k$
(b) $\frac{e^{3+x}}{3} + k$
(c) $e^{x+3} + k$
(d) $3e^{x+3} + k$
›Reveal solutionSolution

$e^3$ is a constant multiplier; $\int e^3 e^x\,dx = e^3 e^x + k = e^{x+3}+k$ .

Treat $e^3$ as a constant and pull it out: …

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CBSE 2025Set E1 markMCQ
$\int 2^{x+1}\,dx =$
(a) $\frac{2^{x+1}}{\log 2} + k$
(b) $2^{x+1} \cdot \log 2 + k$
(c) $(x + 1)2^x + k$
(d) $2^{x+1} + k$
›Reveal solutionSolution

$\int 2^{x+1}\,dx = \frac{2^{x+1}}{\log 2} + k$ .

Write $2^{x+1} = 2\cdot2^{x}$ and use $\int a^x\,dx = \frac{a^x}{\log a} + k$ with $a = 2$ : …

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CBSE 2025Set E1 markMCQ
$\int_{0}^{2} e^x\,dx =$
(a) $e^2$
(b) $e^2 - 2$
(c) $e^2 - 1$
(d) $e - 1$
›Reveal solutionSolution

$\int_0^2 e^x\,dx = [e^x]_0^2 = e^2 - 1$ .

The antiderivative of $e^x$ is $e^x$ . Applying the limits: …

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CBSE 2025Set E1 markMCQ
$\int_{0}^{1} e^x\,dx =$
(a) $e$
(b) $1 - e$
(c) $e - 1$
(d) $0$
›Reveal solutionSolution

$\int_0^1 e^x\,dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ .

The antiderivative of $e^x$ is $e^x$ . Evaluating:

…

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CBSE 2025Set ANNUAL1 markMCQ
$\int e^x . \sec x(1+\tan x)dx$
(a) $e^x\cos x + c$
(b) $e^x\sec x + c$
(c) $e^x\sin x + c$
(d) $e^x\tan x + c$
›Reveal solutionSolution

This fits the standard form $\int e^x[f(x)+f'(x)]dx = e^xf(x)+c$ with $f(x)=\sec x$ .

Expand: $e^x\sec x(1+\tan x) = e^x[\sec x + \sec x\tan x]$ . Since $\dfrac{d}{dx}(\sec x) = \sec x\tan x$ , we have $f(x)=\sec x$ and $f'(x)=\sec x\tan x$ , so the integrand is exactly $e^x[f(x)+f'(x)]$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\int e^x(\sin x + \cos x)\,dx$
(a) $e^x \sin x + C$
(b) $e^x \cos x + C$
(c) $-e^x \sin x + C$
(d) $-e^x \cos x + C$
›Reveal solutionSolution

Use the standard result $\int e^x[f(x)+f'(x)]\,dx = e^xf(x)+C$ .

We use the standard integration formula:

$\int e^x[f(x)+f'(x)]\,dx = e^x f(x) + C$

Here take $f(x) = \sin x$ , so $f'(x) = \cos x$ . The integrand is exactly $e^x(\sin x+\cos x)$ , matching the form $e^x[f(x)+f'(x)]$ with $f(x)=\sin x$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. \int 2^x \cdot 3^x dx equals to ................$
(a) 3^x/ln3 + c
(b) 2^x/ln2 + c
(c) (2^x · 3^x)/(ln2 · ln3) + c
(d) 6^x/ln6 + c
›Reveal solutionSolution

Combine the two exponentials into a single base using $2^x \cdot 3^x = (2\cdot3)^x = 6^x$ , then apply the standard rule $\int a^x\,dx = \dfrac{a^x}{\ln a} + c$ .

Step 1 — combine bases:

$2^x \cdot 3^x = (2 \times 3)^x = 6^x$

…

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CBSE 2024Set ANNUAL1 markMCQ
$\int e^x \sec x (1 + \tan x)\, dx$
(a) $e^x \cos x + c$
(b) $e^x \sec x + c$
(c) $e^x \sin x + c$
(d) $e^x \tan x + c$
›Reveal solutionSolution

$e^x\sec x + c$ — option (b).

Use the standard result $\displaystyle\int e^x\big[f(x)+f'(x)\big]dx = e^xf(x)+c$ .

Take $f(x)=\sec x$ , so $f'(x)=\sec x\tan x$ . Then …

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CBSE 2023Set 65/3/11 markMCQ
$\int 2^{x+2}\, dx$
(a) $2^{x+2} + C$
(b) $2^{x+2} \log 2 + C$
(c) $\frac{2^{x+2}}{\log 2} + C$
(d) $2 \cdot \frac{2^x}{\log 2} + C$
›Reveal solutionSolution
Exponential integrals with base $a$ follow the pattern $\int a^x \, dx = \frac{a^x}{\ln a} + C$ . Rewriting $2^{x+2} = 4 \cdot 2^x$ and integrating gives $\frac{2^{x+2}}{\log 2} + C$ .

The heart of this problem is understanding how exponential functions integrate when the base isn't $e$ . We're comfortable with $\int e^x \, dx = e^x + C$ because $e$ is special—it's its own derivative. But what happens when we have a different base like $2$ ?

The key insight is that any exponential $a^x$ can be rewritten using the natural exponential: $a^x = e^{x \ln a}$ . This connection lets us integrate any exponential function by relating it back to $e^x$ .
$\int a^x \, dx = \frac{a^x}{\ln a} + C$
This formula comes from the chain rule in reverse. When we differentiate $a^x$ , we get $a^x \ln a$ , so when we integrate, we must divide by that same factor $\ln a$ .

Now let's work through the given integral step by step.

Simplify the exponent using exponential laws The expression $2^{x+2}$ can be rewritten as:

$2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$
This factorization pulls out the constant multiplier, making the integral cleaner.

Pull the constant outside the integral

$\int 2^{x+2} \, dx = \int 4 \cdot 2^x \, dx = 4 \int 2^x \, dx$

Apply the exponential integration formula Using our formula with $a = 2$ : $4 \int 2^x \, dx = 4 \cdot \frac{2^x}{\ln 2} + C = \frac{4 \cdot 2^x}{\ln 2} + C$ …
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CBSE 2023Set ANNUAL1 mark
$\displaystyle\int (4e^{3x} + 2)\,dx$
›Reveal solutionSolution

Integrate term by term using $\int e^{ax}dx=\dfrac{1}{a}e^{ax}+C$ .

…

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Q.Find: $\displaystyle\int \frac{e^{4x} - 1}{e^{4x} + 1}\, dx$

Concept understanding — Integration Of Exponential

Integration of Exponential Functions

The idea in one line

The base result

When there is a constant in the exponent

A general base $a^x$

Why the $\log a$ appears

Q.Find: ∫e4x−1e4x+1 dx\displaystyle\int \frac{e^{4x} - 1}{e^{4x} + 1}\, dx∫e4x+1e4x−1​dx

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Q.Find: $\displaystyle\int \frac{e^{4x} - 1}{e^{4x} + 1}\, dx$

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