Q.Integrate the following function: $\frac{1}{\sqrt{7 - 6x - x^2}}$

Showing the 12 most recent of 35 on this concept.

1. 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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2. 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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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2x+5}{\sqrt{7-6x-x^2}} dx = A\sqrt{7-6x-x^2} + B\sin^{-1}\left(\dfrac{x+3}{4}\right) + c$ then the ordered pair $(A,B)=$ (A) $(-2,-1)$ (B) $(2,-1)$ (C) $(-2,1)$ (D) $(2,1)$

   ›Reveal solutionSolution

   Splitting $2x+5$ into a multiple of the derivative of $7-6x-x^2$ plus a constant reduces the integral to a standard $\sqrt{}$ term plus an $\sin^{-1}$ term, giving $(A,B)=(-2,-1)$.

   Concept and Intuition

   For $\int \dfrac{px+q}{\sqrt{ax^2+bx+c}}dx$, always split the numerator as (multiple of the derivative of the quadratic under the root) + (constant), because $\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}$ handles the first part exactly, leaving a pure $1/\sqrt{\text{quadratic}}$ integral for the second part (an inverse-sine form after completing the square).

   Step-by-Step Solution

   1. Let $f(x)=7-6x-x^2$. Then $f'(x)=-6-2x$.
   2. Write $2x+5 = \lambda(-6-2x) + \mu$. Matching coefficients of $x$: $2=-2\lambda\Rightarrow\lambda=-1$. Matching constants: $5=-6\lambda+\mu = 6+\mu\Rightarrow\mu=-1$.
   3. So $2x+5 = -1\cdot(-6-2x) - 1$.
   4. $\displaystyle\int\frac{2x+5}{\sqrt{f(x)}}dx = -1\displaystyle\int\frac{f'(x)}{\sqrt{f(x)}}dx - \int\frac{dx}{\sqrt{f(x)}} = -1\cdot 2\sqrt{f(x)} - \int\frac{dx}{\sqrt{f(x)}}$.
   5. Complete the square: $7-6x-x^2 = -(x^2+6x-7) = -((x+3)^2-16) = 16-(x+3)^2$. …

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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x+1}{\sqrt{x^2+x+1}}\,dx=$ (A) $\dfrac{1}{2}\sqrt{x^2+x+1}+\dfrac{1}{2}\cosh^{-1}\left(\dfrac{x+2}{\sqrt{3}}\right)+c$ (B) $\dfrac{1}{2}\sqrt{x^2+x+1}+\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (C) $\sqrt{x^2+x+1}+\dfrac{2}{\sqrt{3}}\log|x^2+x+1|+c$ (D) $\sqrt{x^2+x+1}+\dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$

   ›Reveal solutionSolution

   Splitting the numerator into a multiple of the derivative of the radicand plus a constant is the standard technique for $\int\dfrac{px+q}{\sqrt{ax^2+bx+c}}dx$; here it gives $\sqrt{x^2+x+1}+\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c$.

   Concept and Intuition

   For $\int\dfrac{\text{linear}}{\sqrt{\text{quadratic}}}dx$, write the linear numerator as $A\cdot(\text{derivative of quadratic})+B$. The $A$-part becomes a simple power-rule integral (since it's exactly $u^{-1/2}du$), and the $B$-part reduces to the standard $\int\dfrac{dx}{\sqrt{x^2+a^2}}=\sinh^{-1}(x/a)+c$ form after completing the square.

   Step-by-Step Solution

   1. Write $x+1=\dfrac12(2x+1)+\dfrac12$.
   2. First part: $\dfrac12\displaystyle\int\dfrac{2x+1}{\sqrt{x^2+x+1}}dx$. Let $u=x^2+x+1$, $du=(2x+1)dx$: this is $\dfrac12\int u^{-1/2}du=\dfrac12\cdot2u^{1/2}=\sqrt{x^2+x+1}$.
   3. Second part: $\dfrac12\displaystyle\int\dfrac{dx}{\sqrt{x^2+x+1}}=\dfrac12\int\dfrac{dx}{\sqrt{(x+1/2)^2+3/4}}$.
   4. This is of the form $\dfrac12\int\dfrac{du}{\sqrt{u^2+a^2}}$ with $u=x+1/2$, $a=\sqrt3/2$, giving $\dfrac12\sinh^{-1}\left(\dfrac{u}{a}\right)=\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)$. …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.For $x > 0$, if $\displaystyle\int \dfrac{1}{x^2 + 5x + 7}\, dx = \dfrac{2}{\sqrt{3}}F(x) + k$ and $F\left(-\dfrac{5}{2}\right) = 0$, then $\sin(F(x)) =$ (A) $\dfrac{2x-5}{\sqrt{3}}$ (B) $\dfrac{2x+5}{2\sqrt{x^2+5x+7}}$ (C) $\dfrac{2\sqrt{x^2+5x+7}}{2x+5}$ (D) $\dfrac{2\sqrt{x^2+5x+7}}{\sqrt{3}}$

   ›Reveal solutionSolution

   Complete the square to integrate the quadratic denominator as a standard arctangent form, identify $F(x)$ from the boundary condition, then convert $\sin(\arctan u)$ into an algebraic expression.

   Concept and Intuition

   Any integral of the form $\int\dfrac{dx}{x^2+px+q}$ reduces to the standard $\int\dfrac{du}{u^2+a^2}=\dfrac1a\arctan\dfrac ua+C$ once the quadratic is written as a completed square. Here the given answer form $\tfrac{2}{\sqrt3}F(x)+k$ tells us $F$ must be exactly that arctangent (up to an additive constant fixed by the given boundary value), after which $\sin(\arctan u)=u/\sqrt{1+u^2}$ finishes the job.

   Step-by-Step Solution

   1. Complete the square: $x^2+5x+7=\left(x+\dfrac52\right)^2+\left(7-\dfrac{25}{4}\right)=\left(x+\dfrac52\right)^2+\dfrac34$.
   2. Standard integral: $\displaystyle\int\dfrac{dx}{(x+5/2)^2+(\sqrt3/2)^2}=\dfrac{1}{\sqrt3/2}\arctan\!\left(\dfrac{x+5/2}{\sqrt3/2}\right)+C=\dfrac{2}{\sqrt3}\arctan\!\left(\dfrac{2x+5}{\sqrt3}\right)+C$.
   3. Matching the given form $\dfrac{2}{\sqrt3}F(x)+k$, take $F(x)=\arctan\!\left(\dfrac{2x+5}{\sqrt3}\right)+c_0$. Since $F(-5/2)=0$: at $x=-5/2$, $\dfrac{2x+5}{\sqrt3}=0$, so $\arctan(0)+c_0=c_0=0$. Thus $F(x)=\arctan\!\left(\dfrac{2x+5}{\sqrt3}\right)$.
   4. Let $u=\dfrac{2x+5}{\sqrt3}$, so $F(x)=\arctan u$ and $\sin(F(x))=\dfrac{u}{\sqrt{1+u^2}}$. …

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8. 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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9. 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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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{x}{\sqrt{x^2 - 2x + 5}}\, dx =$ (A) $\sqrt{x^2-2x+5} + \mathrm{Sinh}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (B) $\dfrac{1}{2}\sqrt{x^2-2x+5} + \mathrm{Sin}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (C) $2\sqrt{x^2-2x+5} + \mathrm{Cosh}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (D) $\sqrt{x^2-2x+5} - \mathrm{Cos}^{-1}\left(\dfrac{x-1}{2}\right) + c$

    ›Reveal solutionSolution

    A rational-times-radical integral of the form $\int\frac{x\,dx}{\sqrt{ax^2+bx+c}}$, split by writing the numerator to match the derivative of the radicand. Answer: $\sqrt{x^2-2x+5}+\mathrm{Sinh}^{-1}\!\left(\frac{x-1}{2}\right)+c$.

    Concept and Intuition

    The standard technique for $\int\dfrac{x\,dx}{\sqrt{x^2+bx+c}}$ is to complete the square in the radicand and write $x$ as (half the derivative of the radicand) plus a constant — this splits the integral into an easy "$u/\sqrt{u^2+a^2}$" piece and a standard inverse hyperbolic-sine piece.

    Step-by-Step Solution

    1. Complete the square: $x^2-2x+5=(x-1)^2+4$.
    2. Let $u=x-1\Rightarrow x=u+1,\ dx=du$. The integral becomes $\displaystyle\int\dfrac{u+1}{\sqrt{u^2+4}}\,du$.
    3. Split: $\displaystyle\int\dfrac{u}{\sqrt{u^2+4}}du+\int\dfrac{du}{\sqrt{u^2+4}}$.
    4. First piece: $\displaystyle\int\dfrac{u\,du}{\sqrt{u^2+4}}=\sqrt{u^2+4}+C_1$ (direct substitution $w=u^2+4$).
    5. Second piece: $\displaystyle\int\dfrac{du}{\sqrt{u^2+4}}=\mathrm{Sinh}^{-1}\!\left(\dfrac{u}{2}\right)+C_2$ (standard form $\int\frac{du}{\sqrt{u^2+a^2}}=\mathrm{Sinh}^{-1}(u/a)$). …

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11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{1}{1+x+x^2}\, dx =$ (A) $\dfrac{2}{\sqrt{3}}\log\left(\dfrac{2x+1+\sqrt{3}}{2x-1-\sqrt{3}}\right) + c$ (B) $\dfrac{1}{\sqrt{3}}\log\left(\dfrac{2x+1-\sqrt{3}}{2x+1+\sqrt{3}}\right) + c$ (C) $\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right) + c$ (D) $\dfrac{2}{\sqrt{5}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{5}}\right) + c$

    ›Reveal solutionSolution

    Completing the square turns the quadratic denominator into a sum of squares, a standard $\int\frac{dx}{u^2+a^2}$ form.

    Concept and Intuition

    Any $\int\dfrac{dx}{ax^2+bx+c}$ with no real roots in the denominator reduces, by completing the square, to the standard arctan integral $\int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c$.

    Step-by-Step Solution

    1. $1+x+x^2=\left(x+\dfrac12\right)^2+1-\dfrac14=\left(x+\dfrac12\right)^2+\dfrac34$.
    2. So the integral is $\displaystyle\int\dfrac{dx}{\left(x+\tfrac12\right)^2+\left(\tfrac{\sqrt3}2\right)^2}$.
    3. Using $\int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c$ with $u=x+\tfrac12$, $a=\dfrac{\sqrt3}2$: $=\dfrac{1}{\sqrt3/2}\tan^{-1}\left(\dfrac{x+1/2}{\sqrt3/2}\right)+c=\dfrac2{\sqrt3}\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c.$ …

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12. 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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