Q.Integrate the following function: $\frac{1}{x (\log x)^m}$, $x>0, m \neq 1$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.$\int(\log x)^m x^n \,dx=$ (A) $\int t^m e^{nt}\,dt,\ t=e^x$ (B) $\int t^m e^{(n+1)t}\,dt,\ t=e^x$ (C) $\int t^m e^{(n+1)t}\,dt,\ x=e^t$ (D) $\int t^m e^{nt}\,dt,\ x=e^t$

   ›Reveal solutionSolution

   The substitution $x=e^t$ (equivalently $t=\log x$) converts a $(\log x)^m x^n$ integral into an exponential-times-power integral in $t$.

   Concept and Intuition

   When an integrand is built from $\log x$ and powers of $x$, setting $x=e^t$ makes $\log x = t$ directly, and turns $x^n\,dx$ into an exponential in $t$ — a very standard substitution for this integral family.

   Step-by-Step Solution

   1. Let $x = e^t$, so $t = \log x$ and $dx = e^t\,dt$.
   2. Then $(\log x)^m = t^m$ and $x^n = (e^t)^n = e^{nt}$.
   3. Substitute into the integral: $\displaystyle\int (\log x)^m x^n\,dx = \int t^m \cdot e^{nt}\cdot e^t\,dt = \int t^m e^{(n+1)t}\,dt$. …

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2. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\int(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{\frac{1}{m}}dx=$ (A) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m+1}{m}}+C$ (B) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$ (C) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6\right)^{\frac{m+1}{m}}+C$ (D) $\dfrac{1}{6(m-1)}\left(2x^{3m}+mx^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$

   ›Reveal solutionSolution

   Recognising that $2x^{3m}+3x^{2m}+6x^m$ equals $x^m$ times the bracket under the $1/m$-power root lets the whole integrand be rewritten as (a constant times) $g^{1/m}g'$ for $g=2x^{3m}+3x^{2m}+6x^m$ — a pure "power rule" integral.

   Concept and Intuition

   Whenever an integrand looks like $g(x)^{1/m}\cdot g'(x)$ (up to a constant factor), the antiderivative is immediately $\dfrac{g^{1/m+1}}{1/m+1}$. The main work here is algebraic: spotting that the "outside" factor $(x^{3m}+x^{2m}+x^m)$ is secretly related to the derivative of $g=2x^{3m}+3x^{2m}+6x^m$, and that the "inside" bracket $(2x^{2m}+3x^m+6)$ is just $g/x^m$.

   Step-by-Step Solution

   1. Let $g=2x^{3m}+3x^{2m}+6x^m$. Factor: $g=x^m(2x^{2m}+3x^m+6)$, so $2x^{2m}+3x^m+6=g/x^m$, and $(2x^{2m}+3x^m+6)^{1/m}=g^{1/m}/x$ (since $(g/x^m)^{1/m}=g^{1/m}/x$).
   2. Differentiate $g$: $g'=6mx^{3m-1}+6mx^{2m-1}+6mx^{m-1}=6mx^{m-1}(x^{2m}+x^m+1)$.
   3. Rewrite the integrand: $(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m}=x^m(x^{2m}+x^m+1)\cdot\dfrac{g^{1/m}}x=x^{m-1}(x^{2m}+x^m+1)\,g^{1/m}$.
   4. From step 2, $x^{m-1}(x^{2m}+x^m+1)=\dfrac{g'}{6m}$. So the integrand equals $\dfrac{g'}{6m}g^{1/m}$. …

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3. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.$\displaystyle\int_{2/e}^{1/e} \dfrac{1}{x(\log x)^{1/3}}\,dx =$ (A) $\dfrac{3}{2}\{1 + (\log(2) - 1)^{2/3}\}$ (B) $1$ (C) $\dfrac{3}{2}\{1 + (\log(2) + 1)^{3/2}\}$ (D) $\dfrac{3}{2}\{1 - (\log(2) - 1)^{2/3}\}$

   ›Reveal solutionSolution

   Substitute $u=\log x$ to turn the integral into $\int u^{-1/3}du$, then evaluate between the transformed limits $u=\log2-1$ and $u=-1$.

   Concept and Intuition

   An integrand of the form $\dfrac{1}{x\cdot g(\log x)}$ always calls for the substitution $u=\log x$, since $du = dx/x$ removes the $x$ and $1/x$ entirely, leaving a pure power of $u$.

   Step-by-Step Solution

   1. Let $u=\log x$, so $du = dx/x$.
   2. Limits: at $x=2/e$, $u=\log(2/e)=\log2-1$. At $x=1/e$, $u=\log(1/e)=-1$.
   3. The integral becomes $\displaystyle\int_{\log2-1}^{-1} u^{-1/3}\,du = \left[\frac{u^{2/3}}{2/3}\right]_{\log2-1}^{-1} = \frac32\Big[u^{2/3}\Big]_{\log2-1}^{-1}$.
   4. Using the real cube root, $(-1)^{2/3} = \big((-1)^{1/3}\big)^2 = (-1)^2 = 1$.
   5. So the value is $\dfrac32\big[1-(\log2-1)^{2/3}\big]$.

   Common Mistakes …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{x^{2022}\left(1+x^{2022}\right)^{1/2022}} = \dfrac{-(1+x^m)^{n/m}}{nx^n} + C$, then $m - n =$ (A) $1$ (B) $2$ (C) $3$ (D) $0$

   ›Reveal solutionSolution

   A standard $\int dx/[x^{k+1}... ]$ trick factoring $x^k$ out of the bracket gives $m=2022$, $n=2021$, hence $m-n=1$.

   Concept and Intuition

   For integrals of the form $\int \dfrac{dx}{x^{k}(1+x^{k})^{1/k}}$, the standard technique is to pull $x^k$ out from inside the bracket, turning it into $(1+x^{-k})$, and then substitute $u=1+x^{-k}$ so that $du$ naturally produces the $x^{-(k+1)}dx$ factor needed.

   Step-by-Step Solution

   1. Write $1+x^{2022} = x^{2022}\left(1+x^{-2022}\right)$, so $(1+x^{2022})^{-1/2022} = x^{-1}(1+x^{-2022})^{-1/2022}$.
   2. The integrand becomes $x^{-2022}\cdot x^{-1}(1+x^{-2022})^{-1/2022} = x^{-2023}(1+x^{-2022})^{-1/2022}$.
   3. Let $u=1+x^{-2022}$, so $du = -2022\,x^{-2023}dx \Rightarrow x^{-2023}dx = -\dfrac{du}{2022}$.
   4. Integral $= \displaystyle\int u^{-1/2022}\left(-\dfrac{du}{2022}\right) = -\dfrac{1}{2022}\cdot\dfrac{u^{2021/2022}}{2021/2022}+C = -\dfrac{u^{2021/2022}}{2021}+C$.
   5. Substitute back: $u^{2021/2022} = (1+x^{-2022})^{2021/2022} = \dfrac{(1+x^{2022})^{2021/2022}}{x^{2021}}$. …

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5. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$ (A) $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$ (B) $\dfrac{2}{(1+\sqrt{x})^{2022}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{\sqrt{x}}{2021}\right] + C$ (C) $\dfrac{2}{(1+\sqrt{x})}\left[\dfrac{(1+\sqrt{x})^{2022}}{2022} - \dfrac{(1+\sqrt{x})^{2021}}{2021}\right] + C$ (D) $\dfrac{1}{(1+\sqrt{x})^2}\left[\dfrac{1}{(1+\sqrt{x})^{1010}} - \dfrac{1}{(1+\sqrt{x})^{1011}}\right] + C$

   ›Reveal solutionSolution

   Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$; back-substituting and factoring reproduces option (A)'s bracketed form.

   Concept and Intuition

   Whenever an integrand is a function purely of $1+\sqrt x$, the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$, $x=(t-1)^2$) turns the messy radical expression into a clean power of $t$, and the pieces of $dx$ that are left over ($2(t-1)\,dt$) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

   Step-by-Step Solution

   1. Let $t=1+\sqrt x$. Then $\sqrt x=t-1$, $x=(t-1)^2$, and $dx=2(t-1)\,dt$.
   2. The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$.
   3. Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$, and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$, combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$. …

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6. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\displaystyle\int_{\log 2}^{\log 3} \dfrac{e^{3x} - 3e^x}{e^{2x} - 1}\,dx =$ (A) $\log\left(\dfrac{2e}{3}\right)$ (B) $\log\left(\dfrac{4e}{3}\right)$ (C) $\dfrac{2e}{3}$ (D) $\dfrac{4e}{3}$

   ›Reveal solutionSolution

   A substitution $t=e^x$ turns this exponential integral into a rational one; the value works out to $\log(2e/3)$.

   Concept and Intuition

   Whenever an integral is built entirely from powers of $e^x$, substituting $t=e^x$ converts it into an algebraic (rational function) integral, which is usually far easier to handle with partial fractions.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^xdx=t\,dx$, i.e. $dx=dt/t$. When $x=\log2$, $t=2$; when $x=\log3$, $t=3$.
   2. Rewrite the integrand: $\dfrac{e^{3x}-3e^x}{e^{2x}-1}=\dfrac{t^3-3t}{t^2-1}$. Multiplying by $dx=dt/t$ gives $\dfrac{t^3-3t}{t(t^2-1)}dt=\dfrac{t^2-3}{t^2-1}dt$.
   3. So the integral becomes $\displaystyle\int_2^3\dfrac{t^2-3}{t^2-1}dt$.
   4. Split: $\dfrac{t^2-3}{t^2-1}=\dfrac{(t^2-1)-2}{t^2-1}=1-\dfrac{2}{t^2-1}$. …

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7. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If $f(x) = \dfrac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$, then $\displaystyle\int x^{n-2} f(x)\, dx =$ (A) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (B) $\dfrac{1}{n-1} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (C) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 + \frac{1}{n}} + C$ (D) $\dfrac{1}{n+1} (1 + nx^n)^{1 + \frac{1}{n}} + C$

   ›Reveal solutionSolution

   A direct substitution $u=1+nx^n$ turns the integral into a simple power rule.

   Concept and Intuition

   The integrand's power of $x$ (namely $x^{n-1}$) is exactly proportional to the derivative of $u=1+nx^n$, which is the classic signal to substitute.

   Step-by-Step Solution

   1. $x^{n-2}f(x)=x^{n-2}\cdot\dfrac{x}{(1+nx^n)^{1/n}}=\dfrac{x^{n-1}}{(1+nx^n)^{1/n}}$.
   2. Let $u=1+nx^n$. Then $du=n\cdot n x^{n-1}dx=n^2x^{n-1}dx$, so $x^{n-1}dx=\dfrac{du}{n^2}$.
   3. Integral becomes $\displaystyle\int u^{-1/n}\cdot\dfrac{du}{n^2}=\dfrac{1}{n^2}\cdot\dfrac{u^{1-1/n}}{1-1/n}+C$.
   4. Simplify: $\dfrac{1}{n^2}\cdot\dfrac{1}{\frac{n-1}{n}}=\dfrac{1}{n^2}\cdot\dfrac{n}{n-1}=\dfrac{1}{n(n-1)}$. …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int_{\log4}^{\log5} \dfrac{e^{2x}+e^x}{e^{2x}-5e^x+6}\,dx=$ (A) $\log\left(\dfrac{64}{9}\right)$ (B) $\log\left(\dfrac{256}{81}\right)$ (C) $\log\left(\dfrac{32}{3}\right)$ (D) $\log\left(\dfrac{128}{27}\right)$

   ›Reveal solutionSolution

   Substituting $t=e^x$ turns the integral into a simple rational-function integral that evaluates to $\log(128/27)$.

   Concept and Intuition

   Whenever an integrand is built entirely out of $e^x$ (here $e^{2x}$ and $e^x$), the substitution $t=e^x$ converts it into a rational function of $t$, which can then be handled by partial fractions — a standard technique for turning transcendental-looking integrals into algebraic ones.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^x\,dx=t\,dx$, i.e. $dx=\dfrac{dt}{t}$. When $x=\log 4$, $t=4$; when $x=\log 5$, $t=5$.
   2. Rewrite the integrand: $e^{2x}+e^x=t^2+t=t(t+1)$, and $e^{2x}-5e^x+6=t^2-5t+6=(t-2)(t-3)$.
   3. So $\displaystyle\int \frac{t(t+1)}{(t-2)(t-3)}\cdot\frac{dt}{t}=\int_4^5 \frac{t+1}{(t-2)(t-3)}\,dt$.
   4. Partial fractions: $\dfrac{t+1}{(t-2)(t-3)}=\dfrac{A}{t-2}+\dfrac{B}{t-3}$. At $t=2$: $A=\dfrac{3}{-1}=-3$. At $t=3$: $B=\dfrac{4}{1}=4$.
   5. So the integral is $\displaystyle\int_4^5\left(\frac{-3}{t-2}+\frac{4}{t-3}\right)dt=\Big[-3\log|t-2|+4\log|t-3|\Big]_4^5$. …

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9. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x-3)^{4/5}(x+1)^{6/5}} =$ (A) $\dfrac{5}{4}\sqrt[5]{\dfrac{x-3}{x+1}} + C$ (B) $\dfrac{5}{4}\left(\dfrac{x+1}{x-3}\right)^{1/5} + C$ (C) $\dfrac{1}{5}\left(\dfrac{x-3}{x+1}\right)^{1/5} + C$ (D) $\dfrac{5}{4}\left(\dfrac{x-3}{x+4}\right)^{4/5} + C$

   ›Reveal solutionSolution

   Splitting off $(x+1)^2$ turns the integrand into a pure power of $t=\frac{x-3}{x+1}$, giving $\frac54\left(\frac{x-3}{x+1}\right)^{1/5}+C$.

   Concept and Intuition

   When an integrand has the form $(x-a)^{p}(x-b)^{q}$ with $p+q$ an integer (here $\frac45+\frac65=2$), factoring out $(x-b)^{p+q}$ and substituting $t=\frac{x-a}{x-b}$ collapses the whole thing to a simple power of $t$ — a standard trick for these "unequal fractional exponent" integrals.

   Step-by-Step Solution

   1. $(x-3)^{4/5}(x+1)^{6/5} = (x+1)^2\left[\dfrac{x-3}{x+1}\right]^{4/5}$ (factoring out $(x+1)^{4/5+6/5}=(x+1)^2$).
   2. So the integrand is $(x+1)^{-2}\left[\dfrac{x-3}{x+1}\right]^{-4/5}$.
   3. Let $t=\dfrac{x-3}{x+1}$. Then $\dfrac{dt}{dx} = \dfrac{(x+1)-(x-3)}{(x+1)^2} = \dfrac{4}{(x+1)^2}$, so $(x+1)^{-2}dx = \dfrac{dt}{4}$.
   4. The integral becomes $\dfrac14\displaystyle\int t^{-4/5}\,dt = \dfrac14\cdot\dfrac{t^{1/5}}{1/5}+C = \dfrac54\,t^{1/5}+C$. …

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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{(x^5+1)^{6/5}} =$ (A) $\dfrac{1}{\sqrt[5]{x^5+1}} + c$ (B) $\dfrac{\sqrt[5]{x^5+1}}{x} + c$ (C) $\dfrac{x}{\sqrt[5]{x^5+1}} + c$ (D) $\sqrt[5]{x^5+1} + c$

    ›Reveal solutionSolution

    Recognizing the integrand as the derivative of $\dfrac{x}{\sqrt[5]{x^5+1}}$ (verified by direct differentiation) gives the antiderivative immediately.

    Concept and Intuition

    For integrals of the form $\int\frac{dx}{(x^n+1)^{(n+1)/n}}$, a useful trick is to guess that the antiderivative looks like $\frac{x}{(x^n+1)^{1/n}}$ (a ratio designed to make the product-rule differentiation collapse nicely), and then verify by differentiating it — if it reproduces the integrand exactly, we're done. This is often faster than a substitution for this particular family.

    Step-by-Step Solution

    1. Guess the antiderivative $g(x) = \dfrac{x}{(x^5+1)^{1/5}} = x(x^5+1)^{-1/5}$.
    2. Differentiate using the product rule: $g'(x) = (x^5+1)^{-1/5} + x\cdot\left(-\dfrac15\right)(x^5+1)^{-6/5}\cdot5x^4$.
    3. Simplify the second term: $x\cdot\left(-\dfrac15\right)(5x^4)(x^5+1)^{-6/5} = -x^5(x^5+1)^{-6/5}$.
    4. So $g'(x) = (x^5+1)^{-1/5} - x^5(x^5+1)^{-6/5}$.
    5. Factor out $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$. …

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11. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{1}{x^5\sqrt[5]{x^5+1}}\,dx=$ (A) $\dfrac{4}{\sqrt[5]{x^5+1}}+c$ (B) $4x^4(x^5+1)^{4/5}+c$ (C) $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$ (D) $-\dfrac{(x^5+1)^{4/5}}{4x^5}+c$

    ›Reveal solutionSolution

    Rewriting the integrand to expose $1+x^{-5}$ as the natural substitution variable solves this cleanly; the answer is $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

    Concept and Intuition

    When an integral mixes a power of $x$ with a root of a polynomial in $x$, factoring out the highest power of $x$ from inside the root often converts the expression into a function of $1/x$ (or $x^{-5}$ here), whose derivative is already present elsewhere in the integrand — a clean substitution.

    Step-by-Step Solution

    1. $(x^5+1)^{-1/5}=\left(x^5\left(1+x^{-5}\right)\right)^{-1/5}=x^{-1}(1+x^{-5})^{-1/5}$.
    2. So the integrand $x^{-5}(x^5+1)^{-1/5}=x^{-6}(1+x^{-5})^{-1/5}$.
    3. Let $t=1+x^{-5}$, so $dt=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dt}{5}$.
    4. Integral $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$. …

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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $n \ge 2$ is a natural number and $0 < \theta < \dfrac{\pi}{2}$, then $\displaystyle\int \dfrac{(\cos^n\theta - \cos\theta)^{1/n}}{\cos^{n+1}\theta}\sin\theta\, d\theta =$ (A) $\dfrac{n}{n-1}\left(\cos^{(1-n)}\theta - 1\right)^2 + c$ (B) $\dfrac{n}{(n+1)(1-n)}\left(\cos^{(1-n)}\theta - 1\right)^{1+\frac{1}{n}} + c$ (C) $\dfrac{1}{n-1}\left(\cos^{(n-1)}\theta - 1\right)^2 + c$ (D) $\dfrac{n}{1-n^2}\left(1 - \cos^{(1-n)}\theta\right)^{(n+1)/n}$

    ›Reveal solutionSolution

    Factor out $\cos^n\theta$ from inside the radical to isolate a clean power of $\cos\theta$, then substitute $w=\cos^{1-n}\theta$ to reduce the whole integral to $\int w^{1/n}dw$.

    Concept and Intuition

    The key move is factoring $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$ so the $n$-th root pulls a clean $\cos\theta$ outside, cancelling nicely against the $\cos^{n+1}\theta$ in the denominator and leaving a single power of $w=\cos^{1-n}\theta$ whose differential exactly matches $\sin\theta\,d\theta/\cos^n\theta$ in the integrand.

    Step-by-Step Solution

    1. Factor: $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$, so

    $(\cos^n\theta-\cos\theta)^{1/n}=\cos\theta\left(1-\cos^{1-n}\theta\right)^{1/n}.$

    2. Divide by $\cos^{n+1}\theta$:

    $\frac{(\cos^n\theta-\cos\theta)^{1/n}}{\cos^{n+1}\theta}=\frac{\left(1-\cos^{1-n}\theta\right)^{1/n}}{\cos^n\theta}.$

    3. Let $w=\cos^{1-n}\theta$. Then $\dfrac{dw}{d\theta}=(1-n)\cos^{-n}\theta\cdot(-\sin\theta)=(n-1)\sin\theta\cos^{-n}\theta$, so $\dfrac{\sin\theta\,d\theta}{\cos^n\theta}=\dfrac{dw}{n-1}$.
    4. The whole integrand times $d\theta$ becomes $(1-w)^{1/n}\cdot\dfrac{dw}{n-1}$ — wait, more directly $w^{1/n}$ is the factor $\left(1-\cos^{1-n}\theta\right)^{1/n}$ once we track $1-\cos^{1-n}\theta$ as the base; carrying the substitution through consistently: …

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