Q.$\int \frac{dx}{2^x + 2^{-x}}$ is equal to: (A) $\tan^{-1}(2^x) + C$ (B) $\tan^{-1}(2^{-x}) + C$ (C) $\frac{\tan^{-1}(2^x)}{\log 2} + C$ (D) $(\log 2)\tan^{-1}(2^x) + C$

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.$\int \frac{dx}{2^x + 2^{-x}}$ is equal to: (A) $\tan^{-1}(2^x) + C$ (B) $\tan^{-1}(2^{-x}) + C$ (C) $\frac{\tan^{-1}(2^x)}{\log 2} + C$ (D) $(\log 2)\tan^{-1}(2^x) + C$

   ›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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2. CBSE 2025Set E1 markMCQ
   Q.$\int \frac{dx}{e^{-x}} =$

   (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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3. CBSE 2025Set E1 markMCQ
   Q.$\int e^3 \cdot e^x\,dx =$

   (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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4. CBSE 2025Set E1 markMCQ
   Q.$\int 2^{x+1}\,dx =$

   (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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5. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{2} e^x\,dx =$

   (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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6. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{1} e^x\,dx =$

   (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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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\int e^x . \sec x(1+\tan x)dx$ is equal to:

   (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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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\int e^x(\sin x + \cos x)\,dx$ equals

   (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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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.∫ 2^x · 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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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.$\int e^x \sec x (1 + \tan x)\, dx$ is equal to:

    (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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11. CBSE 2023Set 65/3/11 markMCQ
    Q.$\int 2^{x+2}\, dx$ is equal to :

    (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.

    1. 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.

    2. Pull the constant outside the integral

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

    3. 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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12. CBSE 2023Set ANNUAL1 mark
    Q.Find the integral: $\displaystyle\int (4e^{3x} + 2)\,dx$.

    ›Reveal solutionSolution

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

    …

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