Q.Integrate the following function: $\frac{x+3}{x^2 - 2x - 5}$

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 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 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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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{1}{x[(\log x)^2+4\log x-1]}dx = A\log\left[\dfrac{\log x+B}{\log x+C}\right]+K$ where K is the constant of integration, then (A) $A=\dfrac{1}{2\sqrt{5}}, B=(2-\sqrt{5}), C=(2+\sqrt{5})$ (B) $A=-\dfrac{1}{2\sqrt{5}}, B=(2-\sqrt{5}), C=(2+\sqrt{5})$ (C) $A=\dfrac{1}{2\sqrt{5}}, B=(2+\sqrt{5}), C=(2-\sqrt{5})$ (D) $A=-\dfrac{1}{2\sqrt{5}}, B=(2+\sqrt{5}), C=(2-\sqrt{5})$

   ›Reveal solutionSolution

   Substituting $t=\log x$ turns the integral into a standard $\int dt/(t^2-a^2)$-type form (after completing the square), directly giving $A,B,C$.

   Concept and Intuition

   Whenever an integral contains $\log x$ repeatedly and $dx/x$ is available (or can be produced), the substitution $t=\log x$ (so $dt=dx/x$) simplifies it into a rational-function integral in $t$, solvable by completing the square and the standard log formula for $\int dt/(t^2-a^2)$.

   Step-by-Step Solution

   1. Let $t=\log x \Rightarrow dt=\dfrac{dx}{x}$. The integral becomes $\displaystyle\int\dfrac{dt}{t^2+4t-1}$.
   2. Complete the square: $t^2+4t-1=(t+2)^2-5$.
   3. $\displaystyle\int\dfrac{dt}{(t+2)^2-5} = \dfrac{1}{2\sqrt5}\log\left|\dfrac{(t+2)-\sqrt5}{(t+2)+\sqrt5}\right|+K$ (standard formula $\int\frac{du}{u^2-a^2}=\frac1{2a}\log\left|\frac{u-a}{u+a}\right|$, with $u=t+2,a=\sqrt5$). …

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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 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{3x+2}{4x^2+4x+5}\, dx = A\log(4x^2+4x+5) + B\tan^{-1}\left(x + \dfrac{1}{2}\right) + C$, then $(A, B) =$ (A) $\left(\dfrac{3}{8}, \dfrac{1}{8}\right)$ (B) $\left(\dfrac{5}{8}, \dfrac{1}{8}\right)$ (C) $\left(-\dfrac{3}{8}, \dfrac{1}{8}\right)$ (D) $\left(-\dfrac{5}{8}, \dfrac{1}{8}\right)$

   ›Reveal solutionSolution

   Splitting the numerator to match the derivative of the denominator plus a constant gives $A=3/8,\,B=1/8$.

   Concept and Intuition

   For $\int\dfrac{px+q}{ax^2+bx+c}dx$, write $px+q$ as a multiple of the denominator's derivative plus a constant remainder, splitting into a log term and an arctan term.

   Step-by-Step Solution

   1. $\dfrac{d}{dx}(4x^2+4x+5)=8x+4$.
   2. Write $3x+2=\dfrac38(8x+4)+k$: $\dfrac38(8x+4)=3x+1.5$, so $k=2-1.5=0.5$.
   3. $\int\dfrac{3x+2}{4x^2+4x+5}dx = \dfrac38\int\dfrac{8x+4}{4x^2+4x+5}dx + \dfrac12\int\dfrac{dx}{4x^2+4x+5}$.
   4. First integral: $\dfrac38\log(4x^2+4x+5)$.
   5. Complete the square: $4x^2+4x+5=4\left[(x+1/2)^2+1\right]$. …

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8. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2x+3}{x(x+1)(x+2)(x+3)+1}\,dx = \dfrac{-1}{ax^2+bx+c} + \alpha$, then value of $a + b + c =$ (A) $3$ (B) $4$ (C) $5$ (D) $6$

   ›Reveal solutionSolution

   Pair up the four linear factors to reveal a perfect-square quadratic denominator, then spot that the numerator is exactly its derivative.

   Concept and Intuition

   Products like $x(x+1)(x+2)(x+3)$ can be regrouped as $[x(x+3)][(x+1)(x+2)]$, both of which share the common quadratic $x^2+3x$ — substituting $t=x^2+3x$ collapses the quartic into a simple quadratic in $t$, and adding 1 often produces a perfect square.

   Step-by-Step Solution

   1. Pair the factors: $x(x+3)=x^2+3x$ and $(x+1)(x+2)=x^2+3x+2$.
   2. Let $t=x^2+3x$. Then $x(x+1)(x+2)(x+3)=t(t+2)=t^2+2t$.
   3. Add the $+1$ from the problem: $t^2+2t+1=(t+1)^2=(x^2+3x+1)^2$.
   4. So denominator $D=(x^2+3x+1)^2$. Notice $\dfrac{d}{dx}(x^2+3x+1)=2x+3$, which is exactly the numerator!
   5. So the integral is $\int\dfrac{u'}{u^2}dx$ where $u=x^2+3x+1$, giving $-\dfrac1u+c=-\dfrac1{x^2+3x+1}+c$. …

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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 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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11. 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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12. 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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