Q.Evaluate: $\int \sqrt{5-2x+x^2}\,dx$

Showing the 12 most recent of 35 on this concept.

1. 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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2. 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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3. 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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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 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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6. 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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7. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{3x^2 + 5x + 4}{\sqrt{x^2 + x + 1}}\, dx = A\sqrt{x^2 + x + 1} + \dfrac{6x}{4}\sqrt{x^2 + x + 1} + B\,\sinh^{-1}\dfrac{2x + 1}{\sqrt{3}} + c$, then $A + 2B =$ (A) $5$ (B) $8$ (C) $\dfrac{31}{8}$ (D) $\dfrac{22}{5}$

   ›Reveal solutionSolution

   Splitting the numerator into a multiple of $(x^2+x+1)$ plus an exact-derivative term, then applying the standard $\int\sqrt{u^2+a^2}\,du$ formula, gives $A=11/4$, $B=9/8$, so $A+2B=5$.

   Concept and Intuition

   For $\int \dfrac{\text{quadratic}}{\sqrt{\text{quadratic}}}dx$, the standard technique writes the numerator as (a multiple of the inner quadratic) + (a multiple of its derivative) + (a constant), reducing the problem to two building blocks: $\int\dfrac{2x+b}{\sqrt{x^2+bx+c}}dx = 2\sqrt{x^2+bx+c}$, and $\int\sqrt{x^2+bx+c}\,dx$ via completing the square into the $\sinh^{-1}$ form.

   Step-by-Step Solution

   1. Write $3x^2+5x+4 = 3(x^2+x+1) + (2x+1)$ (check: $3x^2+3x+3+2x+1=3x^2+5x+4$ ✓).
   2. $\displaystyle\int\dfrac{2x+1}{\sqrt{x^2+x+1}}dx = 2\sqrt{x^2+x+1}$, since $2x+1$ is exactly the derivative of $x^2+x+1$.
   3. For $\displaystyle\int\sqrt{x^2+x+1}\,dx$, complete the square: $x^2+x+1 = \left(x+\tfrac12\right)^2+\tfrac34$. With $u=x+\tfrac12$, $a^2=\tfrac34$: $\int\sqrt{u^2+a^2}\,du = \dfrac{u}{2}\sqrt{u^2+a^2}+\dfrac{a^2}{2}\sinh^{-1}\dfrac{u}{a}+C$.
   4. This gives $\dfrac{x+1/2}{2}\sqrt{x^2+x+1} + \dfrac{3/8}\,\sinh^{-1}\dfrac{2x+1}{\sqrt3} + C = \dfrac{2x+1}{4}\sqrt{x^2+x+1}+\dfrac38\sinh^{-1}\dfrac{2x+1}{\sqrt3}+C$.
   5. Multiply by 3 (from step 1's coefficient): $3\int\sqrt{x^2+x+1}dx = \dfrac{3(2x+1)}{4}\sqrt{x^2+x+1} + \dfrac98\sinh^{-1}\dfrac{2x+1}{\sqrt3}+C$. …

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8. 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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9. 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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10. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{2-\sin x}{2\cos x+3} \, dx =$ (A) $\dfrac{2}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right) - \log\sqrt{2\cos x+3}+c$ (B) $\dfrac{4}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{2}\right) + \log\sqrt{2\cos x+3}+c$ (C) $\dfrac{3}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{2}\right) + \log\sqrt{2\cos x-3}+c$ (D) $\dfrac{1}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{3}\right) - \log\sqrt{2\cos x-3}+c$

    ›Reveal solutionSolution

    Decompose the numerator into a multiple of the denominator's derivative plus a constant; the log part and arctan part combine to option (B).

    Concept and Intuition

    For integrals of the form $\displaystyle\int\dfrac{p+q\sin x}{a+b\cos x}dx$, write the numerator as $\lambda\cdot(\text{derivative of denominator}) + \mu$ (a pure constant), so the integral splits into a straightforward logarithmic piece and a standard $\displaystyle\int\dfrac{dx}{a+b\cos x}$ piece (solved via the Weierstrass/half-angle substitution).

    Step-by-Step Solution

    1. Let $D(x)=2\cos x+3$, so $D'(x)=-2\sin x$.
    2. Write $2-\sin x = \alpha D'(x) + \lambda = -2\alpha\sin x + \lambda$. Matching $\sin x$ coefficients: $-2\alpha=-1\Rightarrow\alpha=\dfrac12$. Matching constants: $\lambda=2$.
    3. So $\displaystyle\int\dfrac{2-\sin x}{2\cos x+3}dx = \dfrac12\int\dfrac{D'(x)}{D(x)}dx + 2\int\dfrac{dx}{2\cos x+3} = \dfrac12\log|2\cos x+3| + 2I$, where $I=\displaystyle\int\dfrac{dx}{2\cos x+3}$.
    4. Standard result (with $a=3,\ b=2$, $a>b$): $I=\dfrac{2}{\sqrt{a^2-b^2}}\tan^{-1}\!\left(\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac x2\right)=\dfrac{2}{\sqrt5}\tan^{-1}\!\left(\dfrac1{\sqrt5}\tan\dfrac x2\right)$ (since $a^2-b^2=5$ and $\sqrt{(a-b)/(a+b)}=\sqrt{1/5}$).
    5. So $2I=\dfrac4{\sqrt5}\tan^{-1}\!\left(\dfrac1{\sqrt5}\tan\dfrac x2\right)$. …

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11. 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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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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