Q.Integrate the following function: $\frac{x^2}{\sqrt{x^6+a^6}}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

   ›Reveal solutionSolution

   The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

   Concept and Intuition

   Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

   Step-by-Step Solution

   1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
   2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
   3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
   4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
   5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$. …

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3. 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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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

   ›Reveal solutionSolution

   Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Concept and Intuition

   When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

   Step-by-Step Solution

   1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
   2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
   3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
   4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$. …

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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 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{x^4 - 1}{x^2\sqrt{x^4 + x^2 + 1}}dx =$ (A) $\frac{2\sqrt{x^4+x^2+1}}{x} + c$ (B) $\frac{\sqrt{x^4+x^2+1}}{x} + c$ (C) $\frac{\sqrt{x^4+x^2+1}}{2x} + c$ (D) $\frac{4\sqrt{x^4+x^2+1}}{x} + c$

   ›Reveal solutionSolution

   Differentiating the candidate $\frac{\sqrt{x^4+x^2+1}}{x}$ reproduces the given integrand exactly, confirming it as the antiderivative.

   Concept and Intuition

   When an integrand looks like it could come from a quotient rule (a square root over a power of $x$), it is often faster to differentiate a plausible candidate of that shape and check, rather than search for a substitution from scratch.

   Step-by-Step Solution

   1. Try $g(x)=\dfrac{\sqrt{x^4+x^2+1}}{x}=\dfrac{N}{x}$ where $N=\sqrt{x^4+x^2+1}$.
   2. $N'=\dfrac{4x^3+2x}{2\sqrt{x^4+x^2+1}}=\dfrac{x(2x^2+1)}{N}$.
   3. Quotient rule: $g'(x)=\dfrac{N'x-N}{x^2}=\dfrac{\frac{x^2(2x^2+1)}{N}-N}{x^2}=\dfrac{x^2(2x^2+1)-N^2}{Nx^2}$.
   4. $N^2=x^4+x^2+1$, so the numerator is $x^2(2x^2+1)-(x^4+x^2+1)=2x^4+x^2-x^4-x^2-1=x^4-1$.
   5. So $g'(x)=\dfrac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$ — exactly the given integrand. …

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7. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$ (A) $\dfrac{-(1+3\sqrt{\tan x})}{(3+\tan^2 x)^3} + C$ (B) $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$ (C) $\dfrac{-(1+\sqrt{\tan x})}{3(1+3\sqrt{\tan x})^2} + C$ (D) $\dfrac{1}{(1+3\sqrt{\tan x})^3} + C$

   ›Reveal solutionSolution

   Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Concept and Intuition

   The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

   Step-by-Step Solution

   1. Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$, matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$.
   2. So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$.
   3. Factor out $\sqrt{\cos x}$: $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$, so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$.
   4. Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$. Let $t=\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$.
   5. Let $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$.
   6. Write $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$. …

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8. 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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9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\sqrt{1 - x^4}}{x^7}\, dx = f(x)\left\{\sqrt{1 - x^4}\right\}^n + c$, then $(f(x))^n =$ ______ (A) $\dfrac{-1}{6x^6}$ (B) $\dfrac{-1}{216x^{18}}$ (C) $\dfrac{1}{36x^{12}}$ (D) $\dfrac{1}{216x^{18}}$

   ›Reveal solutionSolution

   A reduction-formula-style substitution $t=x^{-2}$ turns the integral into a simple power form; matching to the given answer form identifies $f(x)$ and $n$, then $(f(x))^n$ follows directly.

   Concept and Intuition

   Integrals of the form $\int x^m(a+bx^n)^p\,dx$ can often be simplified by substituting $t=x^{-n}$ when $(m+1)/n$ isn't an integer but $(m+1)/n+p$ is — exactly the situation here with $m=-7$, $n=4$, $p=1/2$.

   Step-by-Step Solution

   1. Write the integral as $\displaystyle\int x^{-7}(1-x^4)^{1/2}\,dx$.
   2. Substitute $t = x^{-2}$, so $dt = -2x^{-3}dx \Rightarrow dx = -\dfrac{x^3}{2}dt$, and $x^{-4}=t^2$.
   3. $x^{-7}dx = x^{-7}\cdot\left(-\dfrac{x^3}{2}\right)dt = -\dfrac{x^{-4}}{2}dt = -\dfrac{t^2}{2}dt$.
   4. $1-x^4 = 1-\dfrac{1}{t^2} = \dfrac{t^2-1}{t^2}$, so $\sqrt{1-x^4}=\dfrac{\sqrt{t^2-1}}{t}$ (for $t>0$).
   5. The integral becomes $\displaystyle\int \dfrac{\sqrt{t^2-1}}{t}\cdot\left(-\dfrac{t^2}{2}\right)dt = -\dfrac12\int t\sqrt{t^2-1}\,dt$.
   6. Let $s=t^2-1$, $ds=2t\,dt$: $-\dfrac12\int \sqrt{s}\cdot\dfrac{ds}{2} = -\dfrac14\cdot\dfrac{2}{3}s^{3/2} = -\dfrac16 (t^2-1)^{3/2}+c$. …

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10. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If $\displaystyle\int (x^6 + x^4 + x^2)\sqrt{2x^4 + 3x^2 + 6}\,dx = f(x) + c$, then $f(3) =$ (A) $\dfrac{3}{2}(95)^{3/2}$ (B) $\dfrac{3}{2}(195)^{3/2}$ (C) $\dfrac{3}{2}(265)^{3/2}$ (D) $\dfrac{3}{2}(175)^{3/2}$

    ›Reveal solutionSolution

    Guess the antiderivative in the form $(ax^3+bx)(2x^4+3x^2+6)^{3/2}$, match

    coefficients by differentiating, solve for $a,b$, then evaluate at $x=3$ to get

    $f(3)=\dfrac32(195)^{3/2}$.

    Concept and Intuition

    When an integrand is a polynomial times the square root of another polynomial,

    and the polynomial factor's degree matches what you'd get by differentiating a

    $(\text{poly})\cdot(\text{radicand})^{3/2}$ ansatz, the fastest route is the

    "reverse chain rule with an undetermined polynomial coefficient" method: guess

    the shape of the antiderivative with unknown coefficients, differentiate, and

    match coefficients of like powers of $x$ to solve for them.

    Step-by-Step Solution

    1. Guess $f(x)=(ax^3+bx)\,(2x^4+3x^2+6)^{3/2}$.
    2. Differentiate (product + chain rule), then factor out $(2x^4+3x^2+6)^{1/2}$: $f'(x)=(2x^4+3x^2+6)^{1/2}\Big[(3ax^2+b)(2x^4+3x^2+6)+(ax^3+bx)(12x^3+9x)\Big]$.
    3. Expand the bracket: it collects to $18a\,x^6+(18a+14b)x^4+(18a+12b)x^2+6b$.
    4. Match this to the target $x^6+x^4+x^2$ (constant term $0$): $6b=0\Rightarrow b=0$; then $18a=1\Rightarrow a=\dfrac1{18}$.
    5. Check remaining coefficients with $a=1/18,\ b=0$: $18a+14b=1$ ✓ (matches $x^4$ coefficient $1$), $18a+12b=1$ ✓ (matches $x^2$ coefficient $1$). All consistent.
    6. So $f(x)=\dfrac{x^3}{18}(2x^4+3x^2+6)^{3/2}$. …

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11. 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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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sec^2 x}{(\sec x+\tan x)^{5/2}}\,dx=$ (A) $-\dfrac{(\sec x+\tan x)^{5/2}}{5}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (B) $-\dfrac{(\sec x-\tan x)^{5/2}}{5}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$ (C) $-\dfrac{(\sec x+\tan x)^{3/2}}{3}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (D) $-\dfrac{(\sec x-\tan x)^{3/2}}{3}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$

    ›Reveal solutionSolution

    A substitution $t=\sec x+\tan x$ (which pairs neatly with $\sec x-\tan x=1/t$) reduces this odd-looking integral to a simple power-rule integral, whose answer re-expresses in terms of $\sec x-\tan x$.

    Concept and Intuition

    Whenever $\sec x+\tan x$ appears, remember its reciprocal identity $(\sec x+\tan x)(\sec x-\tan x)=1$, and that $\dfrac{d}{dx}(\sec x+\tan x)=\sec x(\sec x+\tan x)$ — this makes $t=\sec x+\tan x$ a natural substitution whenever $\sec x\,dx$ multiplies a function of $t$.

    Step-by-Step Solution

    1. Let $t=\sec x+\tan x$. Then $dt=\sec x(\sec x+\tan x)\,dx=\sec x\cdot t\,dx$, so $\sec x\,dx=\dfrac{dt}{t}$.
    2. Also $\sec x-\tan x=\dfrac1t$, so $\sec x=\dfrac{t+1/t}{2}=\dfrac{t^2+1}{2t}$.
    3. The integral $\displaystyle\int\frac{\sec^2x}{t^{5/2}}dx=\int\frac{\sec x}{t^{5/2}}\cdot(\sec x\,dx)=\int\frac{\sec x}{t^{5/2}}\cdot\frac{dt}{t}=\int\frac{\sec x}{t^{7/2}}dt$.
    4. Substitute $\sec x=\dfrac{t^2+1}{2t}$: integral $=\displaystyle\int\frac{t^2+1}{2t^{9/2}}dt=\frac12\int\left(t^{-5/2}+t^{-9/2}\right)dt$.
    5. $=\dfrac12\left[-\dfrac{2}{3}t^{-3/2}-\dfrac27 t^{-7/2}\right]+c=-\dfrac{t^{-3/2}}{3}-\dfrac{t^{-7/2}}{7}+c$. …

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