Q.Integrate the following function: $\int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} dx$ equals (A) $10^x - x^{10} + C$ (B) $10^x + x^{10} + C$ (C) $(10^x - x^{10})^{-1} + C$ (D) $\log (10^x + x^{10}) + C$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos x - \sin x}{10 + \sin 2x}\, dx =$ (A) $\dfrac{1}{2}\log(10 + \sin 2x) + c$ (B) $\dfrac{1}{3}\log(10 + \sin 2x) + c$ (C) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$ (D) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}(10 + \sin 2x) + c$

   ›Reveal solutionSolution

   The numerator $\cos x-\sin x$ is exactly $d(\sin x+\cos x)$, and the denominator rewrites in terms of $u=\sin x+\cos x$ via $\sin2x=u^2-1$. That collapses the integral to a standard $\int du/(a^2+u^2)$ arctangent form. Answer: $\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

   Concept and Intuition

   Whenever an integrand contains both $\sin x-\cos x$ (or $\cos x-\sin x$) and $\sin2x$, it is worth trying $u=\sin x+\cos x$ (or $\sin x-\cos x$) as the substitution, because $u^2=1\pm\sin2x$ links the two.

   Step-by-Step Solution

   1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$ — this is exactly the numerator times $dx$.
   2. Also $u^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin2x$, so $\sin2x=u^2-1$.
   3. Denominator: $10+\sin2x=10+u^2-1=9+u^2$.
   4. The integral becomes $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$. …

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2. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{4+5\cos x} =$ (A) $-\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (B) $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (C) $-\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$ (D) $\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$

   ›Reveal solutionSolution

   The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

   Concept and Intuition

   For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$, $dx=\tfrac{2\,dt}{1+t^2}$) always converts the integral into a rational function of $t$ alone.

   Step-by-Step Solution

   1. $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, with $t=\tan(x/2)$.
   2. $4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$.
   3. Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$.
   4. Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$. …

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3. 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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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^3 \tan^{-1} x^4}{1 + x^8}\,dx =$ (A) $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$ (B) $\dfrac{(\tan^{-1}(x^4))^3}{3} + c$ (C) $\dfrac{(\tan^{-1}(x^4))^2}{4} + c$ (D) $\dfrac{(\tan^{-1}(x^4))^2}{2} + c$

   ›Reveal solutionSolution

   A double substitution (first $u=x^4$, then $v=\tan^{-1}u$) turns this into a trivial $\int v\,dv$. Answer: $\dfrac{(\tan^{-1}x^4)^2}{8}+c$.

   Concept and Intuition

   The presence of $x^3\,dx$ alongside $x^4$ inside the arctan and $x^8=(x^4)^2$ in the denominator is a strong signal to substitute $u=x^4$ first. After that, the structure $\tan^{-1}u \cdot \dfrac{du}{1+u^2}$ is exactly of the form "function times its own derivative," solved by a second substitution.

   Step-by-Step Solution

   1. Let $u=x^4$, so $du=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{du}{4}$.
   2. The integral becomes $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$. …

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7. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^4 x}{\left(\sin^2 x + \sin^{-3} x \cos^5 x\right)^{3}}\;dx =$ (A) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-2} + C$ (B) $\dfrac{1}{10}\left(1 + \cot^2 x\right)^{-5} + C$ (C) $\dfrac{1}{10}\left(1 + \cot^5 x\right)^{-2} + C$ (D) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-5} + C$

   ›Reveal solutionSolution

   Factoring $\sin^2x$ out of the denominator turns it into $\sin^2x(1+\cot^5x)$, and the substitution $t=1+\cot^5x$ makes the whole integral a simple power-rule integration, giving $\dfrac{1}{10}(1+\cot^5x)^{-2}+C$.

   Concept and Intuition

   Integrals with mixed powers of $\sin x$ and $\cos x$ in odd/negative combinations often simplify beautifully once you factor out a common power to expose a $(1+\cot^n x)$ or $(1+\tan^n x)$ structure — this is exactly the kind of expression whose derivative (via chain rule) reproduces $\cot^{n-1}x\csc^2x$, matching what's left over in the integrand.

   Step-by-Step Solution

   1. Denominator: $\sin^2x+\sin^{-3}x\cos^5x$. Factor out $\sin^2x$: $=\sin^2x\left[1+\dfrac{\cos^5x}{\sin^5x}\right]=\sin^2x\,(1+\cot^5x)$.
   2. So the full denominator cubed: $\left[\sin^2x(1+\cot^5x)\right]^3=\sin^6x\,(1+\cot^5x)^3$.
   3. Integrand: $\dfrac{\cos^4x}{\sin^6x(1+\cot^5x)^3} = \dfrac{\cos^4x}{\sin^4x}\cdot\dfrac{1}{\sin^2x}\cdot(1+\cot^5x)^{-3}=\cot^4x\,\csc^2x\,(1+\cot^5x)^{-3}$.
   4. Substitute $t=1+\cot^5x$. Then $\dfrac{dt}{dx}=5\cot^4x\cdot(-\csc^2x)=-5\cot^4x\csc^2x$, so $\cot^4x\csc^2x\,dx=-\dfrac{dt}{5}$. …

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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 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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10. 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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11. 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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12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.$\displaystyle\int(\tan^7x+\tan x)\,dx=$ (A) $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$ (B) $\dfrac{\tan^2 x}{6}-\dfrac{\tan^5 x}{4}+\dfrac{\tan^4 x}{2}+c$ (C) $\dfrac{\tan^2 x}{6}(\tan^4 x+3\tan^2 x+4)+c$ (D) $\dfrac{\tan x}{12}(\tan^4 x-3\tan^2 x+6)+c$

    ›Reveal solutionSolution

    Factoring $\tan^7x+\tan x$ using the sum-of-like-terms identity $t^7+t=t(t^6+1)=t(t^2+1)(t^4-t^2+1)$ exposes the $\sec^2x$ needed for a clean $t=\tan x$ substitution. Answer: $\dfrac{\tan^2x}{12}(2\tan^4x-3\tan^2x+6)+c$.

    Concept and Intuition

    Whenever an integrand is built purely from powers of $\tan x$ together with an explicit or hidden $\sec^2x$, substituting $t=\tan x$ turns it into a polynomial integral — the key is recognizing that $t^6+1=(t^2+1)(t^4-t^2+1)$ supplies exactly the $\sec^2x=1+\tan^2x$ factor needed.

    Step-by-Step Solution

    1. $\tan^7x+\tan x=\tan x(\tan^6x+1)=\tan x(\tan^2x+1)(\tan^4x-\tan^2x+1)=\tan x\sec^2x(\tan^4x-\tan^2x+1)$.
    2. Let $t=\tan x$, $dt=\sec^2x\,dx$. Integral becomes $\displaystyle\int t(t^4-t^2+1)dt=\int(t^5-t^3+t)dt$.
    3. $=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$. …

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