Q.Evaluate: $\int \dfrac{dx}{\sqrt{16-9x^2}}$

Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{1}{9\cos^2x - 24\sin x\cos x + 16\sin^2 x}dx =$ (A) $\frac{\cos x}{4(3\cos x - 4\sin x)} + c$ (B) $\frac{\sin x}{4(3\cos x - 4\sin x)} + c$ (C) $\frac{\cos x}{3\cos x - 4\sin x} + c$ (D) $\frac{\sin x}{3\cos x - 4\sin x} + c$

   ›Reveal solutionSolution

   Recognising the denominator as the perfect square $(3\cos x-4\sin x)^2$ and substituting $u=3-4\tan x$ reduces the integral to a simple power rule, giving $\frac{\cos x}{4(3\cos x-4\sin x)}+c$.

   Concept and Intuition

   When a trigonometric denominator has the form $a^2\cos^2x - 2ab\sin x\cos x + b^2\sin^2x$, check whether it is a perfect square $(a\cos x - b\sin x)^2$ first — this instantly turns a messy-looking integral into $\int\sec^2(\cdot)/(\text{linear in }\tan x)^2\,dx$, solvable by substitution.

   Step-by-Step Solution

   1. Observe $9\cos^2x-24\sin x\cos x+16\sin^2x=(3\cos x-4\sin x)^2$ (matches $a^2-2ab+b^2$ with $a=3\cos x,\ b=4\sin x$).
   2. So the integral is $\displaystyle\int\frac{dx}{(3\cos x-4\sin x)^2}$.
   3. Divide numerator and denominator by $\cos^2x$: $=\displaystyle\int\frac{\sec^2x\,dx}{(3-4\tan x)^2}$.
   4. Let $u=3-4\tan x$, so $du=-4\sec^2x\,dx$, i.e. $\sec^2x\,dx=-\tfrac14du$. …

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2. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \frac{dx}{\sqrt{7 - 6x - x^2}} =$ (A) $\mathrm{Sinh}^{-1}\left( \frac{x+3}{4} \right) + c$ (B) $\log\left| \frac{x+3}{4} \right| + c$ (C) $\mathrm{Sin}^{-1}\left( \frac{x+3}{4} \right) + c$ (D) $\frac{1}{2}\mathrm{Sin}^{-1}\left( \frac{x+3}{4} \right) + c$

   ›Reveal solutionSolution

   Completing the square under the root reveals the standard form $\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}(u/a)$. Answer: $\sin^{-1}\left(\frac{x+3}{4}\right)+c$.

   Concept and Intuition

   A quadratic under a square root, when completed to the square, reveals which standard integral form applies: $a^2-(x-h)^2$ gives an arcsine, while $(x-h)^2+a^2$ or $(x-h)^2-a^2$ give hyperbolic-inverse/log forms.

   Step-by-Step Solution

   1. $7-6x-x^2 = -(x^2+6x-7) = -[(x+3)^2-9-7] = -(x+3)^2+16 = 16-(x+3)^2$.
   2. So the integral is $\displaystyle\int\frac{dx}{\sqrt{4^2-(x+3)^2}}$.
   3. Using $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac ua\right)+c$ with $u=x+3,\ a=4$: …

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3. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(2ax+x^2)^{3/2}}=$ (A) $\dfrac{1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$ (B) $\dfrac{1}{a^2}\left(\dfrac{x-a}{\sqrt{2ax+x^2}}\right)+C$ (C) $\dfrac{-1}{a^2}\left(\dfrac{x-a}{\sqrt{2ax+x^2}}\right)+C$ (D) $\dfrac{-1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$

   ›Reveal solutionSolution

   Completing the square converts the integral into the standard form $\int du/(u^2-a^2)^{3/2}$, which has a known closed form; back-substituting gives option (D).

   Concept and Intuition

   Many integrals of the form $\int dx/(\text{quadratic})^{3/2}$ become standard once the quadratic is completed to a perfect-square-minus-constant form; the resulting substitution $u=x+a$ reduces it to a memorized/derivable antiderivative.

   Step-by-Step Solution

   1. $2ax+x^2=x^2+2ax+a^2-a^2=(x+a)^2-a^2$. Let $u=x+a$, $du=dx$.
   2. The integral becomes $\displaystyle\int\frac{du}{(u^2-a^2)^{3/2}}$.
   3. Standard result (verifiable by differentiation): $\displaystyle\int\frac{du}{(u^2-a^2)^{3/2}}=\frac{-u}{a^2\sqrt{u^2-a^2}}+C$. Check: $\dfrac{d}{du}\left[\dfrac{-u}{a^2\sqrt{u^2-a^2}}\right]=\dfrac{1}{(u^2-a^2)^{3/2}}$ ✓. …

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4. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.$\int \dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx=$ (A) $\dfrac{1}{2x^2}\sqrt{2x^4+2x^2+1}+C$ (B) $\dfrac{1}{2x^2}\sqrt{2x^4-2x^2+1}+C$ (C) $\dfrac{1}{2x^2}\sqrt{4x^4-2x^2+1}+C$ (D) $\dfrac{1}{2x^2}\sqrt{4x^4+2x^2+1}+C$

   ›Reveal solutionSolution

   Verify by differentiation: $F(x)=\dfrac{1}{2x^2}\sqrt{2x^4-2x^2+1}$ differentiates back to the given integrand, so this is the antiderivative.

   Concept and Intuition

   When an integrand looks like it could come from differentiating a quotient of the form $\dfrac{\sqrt{\text{quartic}}}{x^2}$, the fastest rigorous route (especially under exam time pressure) is to differentiate the most plausible option and check it reproduces the integrand exactly, rather than deriving the substitution from scratch.

   Step-by-Step Solution

   1. Let $u=\sqrt{2x^4-2x^2+1}$ and $F=\dfrac{u}{2x^2}$.
   2. $u'=\dfrac{8x^3-4x}{2u}=\dfrac{4x^3-2x}{u}$.
   3. Quotient rule: $F'=\dfrac{u' \cdot 2x^2 - u\cdot 4x}{4x^4}=\dfrac{\frac{(4x^3-2x)}{u}\cdot 2x^2-4xu}{4x^4}$.
   4. Multiply numerator and denominator by $u$: $F'=\dfrac{2x^2(4x^3-2x)-4xu^2}{4x^4 u}=\dfrac{8x^5-4x^3-4x(2x^4-2x^2+1)}{4x^4u}$.
   5. Expand the numerator: $8x^5-4x^3-8x^5+8x^3-4x=4x^3-4x$. …

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

   ›Reveal solutionSolution

   Verifying the antiderivative by differentiating each candidate is faster and safer than guessing a substitution; the derivative of option (D) reproduces the integrand exactly. Answer: option (D).

   Concept and Intuition

   When an integral has an awkward-looking algebraic form under a square root, and the options are all algebraic expressions (not transcendental), the fastest rigorous check is to differentiate the candidate answers and see which one reproduces the integrand — this avoids errors in choosing a substitution.

   Step-by-Step Solution

   1. Let $N(x) = 1-2x^2+2x^4$ (note this equals $2x^4-2x^2+1$, the expression under the root in the integrand).
   2. Try $f(x) = \dfrac{N^{1/2}}{2x^2} = \dfrac12 N^{1/2}x^{-2}$.
   3. Differentiate: $f'(x) = \dfrac{N'}{4N^{1/2}x^2} - \dfrac{N^{1/2}}{x^3}$, where $N'=-4x+8x^3=4x(2x^2-1)$.
   4. First term becomes $\dfrac{(2x^2-1)}{xN^{1/2}}$. Combine both terms over the common denominator $x^3N^{1/2}$:

   $f'(x)=\frac{(2x^2-1)x^2 - N}{x^3N^{1/2}}.$

   5. Numerator: $(2x^2-1)x^2-N = (2x^4-x^2)-(1-2x^2+2x^4) = x^2-1$. …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(1 + \sqrt{x})\sqrt{x - x^2}} =$ (A) $-2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$ (B) $-\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (C) $-2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (D) $2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$

   ›Reveal solutionSolution

   Two successive substitutions ($t=\sqrt x$, then $w=\sqrt{(1-t)/(1+t)}$) collapse the integral to $\int -2\,dw$; the answer is $-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

   Concept and Intuition

   The presence of $\sqrt x$ inside $(1+\sqrt x)$ and inside $\sqrt{x-x^2}=\sqrt x\sqrt{1-x}$ both suggest first substituting $t=\sqrt x$. What remains — a rational-times-square-root expression in $t$ symmetric under $t\to -t$ in a $(1\pm t)$ sense — is the classic cue for the substitution $w^2=\dfrac{1-t}{1+t}$, which rationalizes everything at once.

   Step-by-Step Solution

   1. Note $x-x^2=x(1-x)$, so $\sqrt{x-x^2}=\sqrt x\,\sqrt{1-x}$, and the integral is

   $\int\frac{dx}{(1+\sqrt x)\sqrt x\sqrt{1-x}}.$

   2. Let $t=\sqrt x$, so $x=t^2$, $dx=2t\,dt$:

   $\int\frac{2t\,dt}{(1+t)\,t\,\sqrt{1-t^2}}=\int\frac{2\,dt}{(1+t)\sqrt{(1-t)(1+t)}}=\int\frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}.$

   3. Let $w=\sqrt{\dfrac{1-t}{1+t}}$, so $t=\dfrac{1-w^2}{1+w^2}$ and $dt=\dfrac{-4w}{(1+w^2)^2}\,dw$. One finds

   $1+t=\frac{2}{1+w^2},\qquad 1-t=\frac{2w^2}{1+w^2},$

   so

   $(1+t)^{3/2}(1-t)^{1/2}=\frac{4w}{(1+w^2)^2}.$

   4. Substituting, …

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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x^2+x+1}\,dx$ (A) $\dfrac{(2x+1)}{4}\sqrt{x^2+x+1}+\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (B) $\dfrac{x+1}{4}\sqrt{x^2+x+1}+\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (C) $\dfrac{x+1}{4}\sqrt{x^2+x+1}-\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (D) $\dfrac{(2x+1)}{4}\sqrt{x^2+x+1}-\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$

   ›Reveal solutionSolution

   Completing the square turns $\sqrt{x^2+x+1}$ into the standard $\sqrt{u^2+a^2}$ form, whose known integral gives option (A).

   Concept and Intuition

   The standard result $\displaystyle\int\sqrt{u^2+a^2}\,du=\frac u2\sqrt{u^2+a^2}+\frac{a^2}2\,\mathrm{Sinh}^{-1}\!\left(\frac ua\right)+c$ applies to any quadratic under a square root once it's written as a perfect square plus a constant.

   Step-by-Step Solution

   1. Complete the square: $x^2+x+1=\left(x+\dfrac12\right)^2+\dfrac34$. Let $u=x+\dfrac12$, $a^2=\dfrac34$ (so $a=\dfrac{\sqrt3}2$).
   2. Apply the formula: $\displaystyle\int\sqrt{u^2+a^2}\,du=\frac u2\sqrt{u^2+a^2}+\frac{a^2}2\mathrm{Sinh}^{-1}\!\left(\frac ua\right)+c$.
   3. $\dfrac u2 = \dfrac{x+\frac12}{2}=\dfrac{2x+1}{4}$, and $\sqrt{u^2+a^2}=\sqrt{x^2+x+1}$.
   4. $\dfrac{a^2}2=\dfrac{3/4}{2}=\dfrac38$, and $\dfrac ua = \dfrac{x+\frac12}{\sqrt3/2}=\dfrac{2x+1}{\sqrt3}$. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{1 - \sin^4 x} = A\tan x + B\tan^{-1}(\sqrt{2}\tan x) + C$, then $A^2 - B^2 =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{1}{8}$

   ›Reveal solutionSolution

   Factoring $1-\sin^4x$ and substituting $t=\tan x$ reduces the integral to a standard rational form, giving $A=\frac12$, $B=\frac{1}{2\sqrt2}$, and $A^2-B^2=\frac18$.

   Concept and Intuition

   $1-\sin^4x$ is a difference of squares: $(1-\sin^2x)(1+\sin^2x)=\cos^2x(1+\sin^2x)$. Pulling out one factor of $\cos^2x$ to pair with $dx$ suggests the substitution $t=\tan x$, turning the whole integral into a rational function of $t$ that splits into a linear term plus an arctan term.

   Step-by-Step Solution

   1. Factor: $1-\sin^4x = (1-\sin^2x)(1+\sin^2x) = \cos^2x(1+\sin^2x)$.
   2. So the integrand is $\dfrac{1}{\cos^2x(1+\sin^2x)} = \dfrac{\sec^2x}{1+\sin^2x}$.
   3. Substitute $t=\tan x$, $dt=\sec^2x\,dx$, and $\sin^2x = \dfrac{t^2}{1+t^2}$, so $1+\sin^2x = \dfrac{1+2t^2}{1+t^2}$.
   4. The integral becomes $\displaystyle\int \frac{1+t^2}{1+2t^2}\,dt$.
   5. Simplify: $\dfrac{1+t^2}{1+2t^2} = \dfrac12\cdot\dfrac{(1+2t^2)+1}{1+2t^2} = \dfrac12 + \dfrac{1}{2(1+2t^2)}$.
   6. Integrate: $\displaystyle\int\left[\frac12+\frac{1}{2(1+2t^2)}\right]dt = \frac{t}{2} + \frac12\cdot\frac{1}{\sqrt2}\tan^{-1}(\sqrt2 t) + C = \frac{t}{2}+\frac{1}{2\sqrt2}\tan^{-1}(\sqrt2 t)+C$. …

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9. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\displaystyle\int \left(\sqrt{x + \sqrt{12x - 36}} + \sqrt{x - \sqrt{12x - 36}}\right) dx =$ (A) $2\sqrt{3}x + C, \ \forall x$ (B) $\dfrac{4(x - 3)^{3/2}}{3} + C, \ \forall x$ (C) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & x > 6 \\ 2\sqrt{3}x + C, & 3 \leq x \leq 6 \end{cases}$ (D) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & 3 \leq x \leq 6 \\ 2\sqrt{3}x + C, & x > 6 \end{cases}$

   ›Reveal solutionSolution

   Substituting $t=\sqrt{x-3}$ turns both nested radicals into perfect squares $(t\pm\sqrt3)^2$; the integrand collapses to a constant $2\sqrt3$ on $[3,6]$ and to $2\sqrt{x-3}$ for $x>6$, giving the piecewise antiderivative in (C).

   Concept and Intuition

   Nested radicals of the form $\sqrt{x\pm\sqrt{\text{linear in }x}}$ often simplify to perfect squares under a substitution that removes the inner square root. Here the key is recognizing $12x-36$ is a perfect multiple of $(x-3)$.

   Step-by-Step Solution

   1. Let $t=\sqrt{x-3}\ge0$ (valid for $x\ge3$), so $x=t^2+3$ and $12x-36=12t^2\Rightarrow \sqrt{12x-36}=2\sqrt3\,t$.
   2. $x+\sqrt{12x-36} = t^2+3+2\sqrt3\,t = (t+\sqrt3)^2 \Rightarrow \sqrt{x+\sqrt{12x-36}} = t+\sqrt3$ (always non-negative).
   3. $x-\sqrt{12x-36} = t^2+3-2\sqrt3\,t=(t-\sqrt3)^2 \Rightarrow \sqrt{x-\sqrt{12x-36}} = |t-\sqrt3|$.
   4. Sum $= (t+\sqrt3)+|t-\sqrt3|$. If $t\ge\sqrt3$ (i.e. $x-3\ge3 \Rightarrow x\ge6$): sum $=2t=2\sqrt{x-3}$. If $0\le t<\sqrt3$ (i.e. $3\le x<6$): sum $=(t+\sqrt3)+(\sqrt3-t)=2\sqrt3$. …

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

    ›Reveal solutionSolution

    Complete the square in the denominator and apply the standard $\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}(x/a)$ form. Answer: option (B).

    Concept and Intuition

    Any irreducible quadratic $ax^2+bx+c$ in a denominator under a simple rational integrand can be handled by completing the square to reduce it to the standard $u^2+a^2$ form, whose antiderivative is a scaled arctangent.

    Step-by-Step Solution

    1. Complete the square: $1+x+x^2 = \left(x+\dfrac12\right)^2 + \dfrac34 = \left(x+\dfrac12\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2$.
    2. So the integral is $\displaystyle 2\int \frac{dx}{(x+\frac12)^2+(\frac{\sqrt3}{2})^2}$.
    3. Using $\int \frac{du}{u^2+a^2}=\frac1a\tan^{-1}(u/a)$ with $u=x+\frac12$, $a=\frac{\sqrt3}{2}$:

    $2\cdot\frac{1}{\sqrt3/2}\tan^{-1}\!\left(\frac{x+1/2}{\sqrt3/2}\right) = \frac{4}{\sqrt3}\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right)+c.$

    Common Mistakes …

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11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sqrt{x-2}}{2x+4}\, dx =$ (A) $\sqrt{x-2} - \dfrac{1}{2}\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (B) $\sqrt{x-2} - 2\,\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (C) $\sqrt{x-2} + 2\,\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (D) $\sqrt{x-2} + \dfrac{1}{2}\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$

    ›Reveal solutionSolution

    A rationalizing substitution $x-2=t^2$ converts the integral into a simple $\int\left(1-\dfrac{4}{t^2+4}\right)dt$, giving $\sqrt{x-2}-2\,\mathrm{Tan}^{-1}\!\left(\dfrac{\sqrt{x-2}}{2}\right)+c$.

    Concept and Intuition

    Whenever an integral has a single square root of a linear expression (here $\sqrt{x-2}$), the substitution $x-2=t^2$ (so $t=\sqrt{x-2}$) removes the square root entirely and typically converts the integral into a rational function of $t$, which is then handled by the standard $\int\dfrac{dt}{t^2+a^2}$ arctan formula.

    Step-by-Step Solution

    1. Simplify the denominator first: $2x+4=2(x+2)$, so the integral is $\dfrac12\displaystyle\int\frac{\sqrt{x-2}}{x+2}dx$.
    2. Substitute $x-2=t^2 \Rightarrow x = t^2+2$, $dx=2t\,dt$, and $x+2 = t^2+4$.
    3. The integral becomes

    $\frac12\int\frac{t}{t^2+4}\cdot 2t\,dt = \int\frac{t^2}{t^2+4}dt.$

    4. Split the rational function: $\dfrac{t^2}{t^2+4}=1-\dfrac{4}{t^2+4}$.
    5. Integrate termwise: $\displaystyle\int\left(1-\frac{4}{t^2+4}\right)dt = t - 4\cdot\frac12\,\mathrm{Tan}^{-1}\!\left(\frac t2\right)+c = t-2\,\mathrm{Tan}^{-1}\!\left(\frac t2\right)+c$.
    6. Substitute back $t=\sqrt{x-2}$: …

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12. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\, dx =$ (A) $\log|x| - \dfrac{1}{4x^4} + C$ (B) $\log|x| + \dfrac{1}{4x^4} + C$ (C) $\log|x| - \dfrac{4}{x^4} + C$ (D) $\log|x| + \dfrac{4}{x^4} + C$

    ›Reveal solutionSolution

    The expression under the square root is a perfect square, simplifying the integral to a sum of two elementary power terms.

    Concept and Intuition

    Recognize $x^4+x^{-4}+2$ as $(x^2+x^{-2})^2$ using the identity $a^2+b^2+2ab=(a+b)^2$ with $a=x^2,b=x^{-2}$.

    Step-by-Step Solution

    1. $(x^2+x^{-2})^2 = x^4+2+x^{-4}$ — matches the expression under the square root exactly.
    2. $\sqrt{x^4+x^{-4}+2} = |x^2+x^{-2}| = x^2+\dfrac{1}{x^2}$ (always positive for real $x\ne0$).
    3. Divide by $x^3$: $\dfrac{x^2+1/x^2}{x^3} = \dfrac1x+\dfrac{1}{x^5}$. …

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