Q.$\int \frac{dx}{2 - 3x} =$

Showing the 12 most recent of 25 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{dx}{x - 1} =$

   (a) $\log|x + 1| + k$

   (b) $-\log|x + 1| + k$

   (c) $\log|x - 1| + k$

   (d) $\log x + k$

   ›Reveal solutionSolution

   Standard log integral: $\int\tfrac{dx}{x-1}=\log|x-1|+k$.

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.∫ tan x dx equals

   (a) log|sin x| + C

   (b) log|cos x| + C

   (c) log|sec x| + C

   (d) log|cot x| + C

   ›Reveal solutionSolution

   Write $\tan x$ as $\sin x/\cos x$ and integrate by substitution $u=\cos x$.

   $\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$

   Let $u=\cos x$, so $du=-\sin x\,dx$, i.e. $\sin x\,dx = -du$.

   …

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3. CBSE 2025Set E1 markMCQ
   Q.$\int \frac{dx}{x\log x} =$

   (a) $\log x + k$

   (b) $(\log x)^2 + k$

   (c) $\log(\log x) + k$

   (d) $\frac{1}{\log x} + k$

   ›Reveal solutionSolution

   With $t = \log x$, the integral becomes $\int \frac{dt}{t} = \log(\log x)+k$.

   Let $t = \log x$, so $dt = \frac{1}{x}\,dx$. Then …

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4. CBSE 2025Set E1 markMCQ
   Q.$\int \frac{x - 3}{x^2 - 9}\,dx =$

   (a) $\log(x - 3) + k$

   (b) $\log(x + 3) + k$

   (c) $-\frac{1}{(x + 3)^2} + k$

   (d) $\frac{x^2}{2} - 3x + k$

   ›Reveal solutionSolution

   $\frac{x-3}{x^2-9} = \frac{1}{x+3}$, so the integral is $\log(x+3)+k$.

   Factor the denominator using $x^2 - 9 = (x-3)(x+3)$:

   $\frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3}.$ …

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5. CBSE 2024Set D1 markMCQ
   Q.$\int \sec x\,dx =$

   (a) $\log|\sec x + \tan x| + c$

   (b) $\log|\sec x - \tan x| + c$

   (c) $\log \sec x + c$

   (d) $\tan^5 x + c$

   ›Reveal solutionSolution

   $\int\sec x\,dx=\log|\sec x+\tan x|+c$.

   Multiplying numerator and denominator by $(\sec x+\tan x)$ and substituting $t=\sec x+\tan x$ gives the standard result:

   …

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6. CBSE 2024Set D1 markMCQ
   Q.$\int \frac{dx}{2 - 3x} =$

   (a) $-3\log|2 - 3x| + c$

   (b) $-\frac{1}{3}\log|2 - 3x| + c$

   (c) $-\log|2 - 3x| + c$

   (d) $2\tan^{-1} x^4 + c$

   ›Reveal solutionSolution

   $\int \frac{dx}{a+bx} = \frac{1}{b}\log|a+bx| + c$; here $b = -3$.

   Let $u = 2 - 3x$, so $du = -3\,dx$, i.e. $dx = -\frac{1}{3}du$. Then …

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7. CBSE 2024Set ANNUAL1 mark
   Q.Evaluate $\displaystyle\int \dfrac{\sin x}{1 + \cos x}\,dx$.

   ›Reveal solutionSolution

   Substitute $u = 1+\cos x$; then $du = -\sin x\,dx$, turning the integral into a simple $\dfrac{du}{u}$ form.

   $\displaystyle\int \dfrac{\sin x}{1+\cos x}\, dx$

   Let $u = 1 + \cos x$, so $du = -\sin x\, dx$, i.e. $\sin x\, dx = -du$.

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\int_2^3 \frac{1}{x} \,dx =$

   (a) $\frac{2}{3}\log x$

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

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

   (d) none of these

   ›Reveal solutionSolution

   Integrate 1/x to get log x, then apply the limits and combine using log rules.

   …

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9. CBSE 2024Set ANNUAL1 mark
   Q.Evaluate : $\displaystyle\int \dfrac{dx}{3x+2}$.

   ›Reveal solutionSolution

   Direct substitution $u=3x+2$.

   $\int\dfrac{dx}{3x+2}$

   Let $u=3x+2 \implies du=3\,dx \implies dx=\dfrac{du}{3}$. …

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10. CBSE 2023Set E1 markMCQ
    Q.$\displaystyle\int\dfrac{x+2}{x^{2}-4}\,dx =$

    (a) $\log|x+2|+k$

    (b) $\log|x^{2}-4|+k$

    (c) $\log|x-2|+k$

    (d) $\log\left|\dfrac{x+2}{x-2}\right|+k$

    ›Reveal solutionSolution

    Cancelling $(x+2)$ reduces the integrand to $\dfrac{1}{x-2}$, whose integral is $\log|x-2|+k$.

    Factor the denominator: $x^{2}-4=(x-2)(x+2)$.

    …

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11. CBSE 2023Set E1 markMCQ
    Q.$\displaystyle\int\dfrac{3\cos x-2\sin x}{2\cos x+3\sin x}\,dx =$

    (a) $2\cos x+3\sin x+k$

    (b) $\log|2\cos x+3\sin x|+k$

    (c) $\tan^{-1}\left(3\sin\dfrac{x}{2}\right)+k$

    (d) $2\tan\dfrac{x}{2}+k$

    ›Reveal solutionSolution

    Since $\dfrac{d}{dx}(2\cos x+3\sin x)=3\cos x-2\sin x$, the integral is of the form $\int\tfrac{f'}{f}=\log|f|$.

    Let $f(x)=2\cos x+3\sin x$. Then $f'(x)=-2\sin x+3\cos x=3\cos x-2\sin x$, which is the numerator.

    Using $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+k$: …

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12. CBSE 2023Set E1 markMCQ
    Q.$\displaystyle\int\dfrac{3x^{2}+2}{x^{3}+2x}\,dx =$

    (a) $\sin^{-1}(x^{3}+2x)+k$

    (b) $\tan^{-1}(3x^{2}+2)+k$

    (c) $\log|3x^{2}+2|+k$

    (d) $\log|x^{3}+2x|+k$

    ›Reveal solutionSolution

    Since $\dfrac{d}{dx}(x^{3}+2x)=3x^{2}+2$, the integral is $\int\tfrac{f'}{f}=\log|f|$.

    Let $f(x)=x^{3}+2x$. Then $f'(x)=3x^{2}+2$, exactly the numerator.

    Using $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+k$: …

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