Q.Integrate the following function: $\frac{5x-2}{1+2x+3x^2}$

Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{(3x-2)\tan(\sqrt{9x^2-12x+1})}{\sqrt{9x^2-12x+1}}dx =$ (A) $\frac{1}{3}\sec^2\sqrt{9x^2-12x+1} + c$ (B) $\frac{1}{3}\sec^2 x + c$ (C) $\frac{1}{2}\log\left|\sec\sqrt{9x^2-12x+1}\right| + c$ (D) $\frac{1}{3}\log\left|\sec\sqrt{9x^2-12x+1}\right| + c$

   ›Reveal solutionSolution

   A double substitution — first $u=9x^2-12x+1$, then $w=\sqrt u$ — turns the integral into a plain $\int\tan w\,dw$, giving $\frac13\log|\sec\sqrt{9x^2-12x+1}|+c$.

   Concept and Intuition

   When $(3x-2)$ (half the derivative of $9x^2-12x+1$) sits outside a function of $\sqrt{9x^2-12x+1}$, a chained substitution — first for the quadratic, then for its square root — collapses the whole expression to a single-variable standard integral.

   Step-by-Step Solution

   1. Let $u=9x^2-12x+1$. Then $du=(18x-12)\,dx=6(3x-2)\,dx$, so $(3x-2)\,dx=\dfrac{du}{6}$.
   2. Integral becomes $\displaystyle\int\frac{\tan\sqrt u}{\sqrt u}\cdot\frac{du}{6}=\frac16\int\frac{\tan\sqrt u}{\sqrt u}\,du$.
   3. Let $w=\sqrt u$, so $dw=\dfrac{du}{2\sqrt u}$, i.e. $\dfrac{du}{\sqrt u}=2\,dw$.
   4. Integral becomes $\dfrac16\int\tan w\cdot2\,dw=\dfrac13\int\tan w\,dw=\dfrac13\left(-\log|\cos w|\right)+c=\dfrac13\log|\sec w|+c$. …

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2. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x-1)\sqrt{x+2}} =$ (A) $\dfrac{2}{\sqrt{3}}\log\left|\dfrac{\sqrt{(x+2)} + \sqrt{3}}{\sqrt{(x+2)} - \sqrt{3}}\right| + c$ (B) $\dfrac{-1}{\sqrt{3}}\log\left|\dfrac{\sqrt{(x+2)} - \sqrt{3}}{\sqrt{(x+2)} + \sqrt{3}}\right| + c$ (C) $\dfrac{1}{\sqrt{3}}\log\left|\dfrac{\sqrt{(x+2)} + \sqrt{3}}{\sqrt{(x+2)} - \sqrt{3}}\right| + c$ (D) $\dfrac{1}{\sqrt{3}}\log\left|\dfrac{\sqrt{(x+2)} - \sqrt{3}}{\sqrt{(x+2)} + \sqrt{3}}\right| + c$

   ›Reveal solutionSolution

   The substitution $t=\sqrt{x+2}$ turns the integral into the standard $\int\frac{dt}{t^2-a^2}$ form.

   Concept and Intuition

   Whenever an integrand has $\sqrt{\text{linear}}$ and a polynomial in $x$, substituting $t=\sqrt{\text{linear expression}}$ often rationalizes everything.

   Step-by-Step Solution

   1. Let $t=\sqrt{x+2}\Rightarrow x=t^2-2,\ dx=2t\,dt$.
   2. $x-1=t^2-2-1=t^2-3$.
   3. Integral becomes $\displaystyle\int\dfrac{2t\,dt}{(t^2-3)\cdot t}=\int\dfrac{2\,dt}{t^2-3}$.
   4. Using $\displaystyle\int\dfrac{dt}{t^2-a^2}=\dfrac{1}{2a}\log\left|\dfrac{t-a}{t+a}\right|+C$ with $a=\sqrt3$: $\int\dfrac{2\,dt}{t^2-3}=2\cdot\dfrac{1}{2\sqrt3}\log\left|\dfrac{t-\sqrt3}{t+\sqrt3}\right|+C=\dfrac{1}{\sqrt3}\log\left|\dfrac{t-\sqrt3}{t+\sqrt3}\right|+C$.
   5. Substitute back $t=\sqrt{x+2}$: $\dfrac{1}{\sqrt3}\log\left|\dfrac{\sqrt{x+2}-\sqrt3}{\sqrt{x+2}+\sqrt3}\right|+c$. …

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3. 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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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 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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6. 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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7. 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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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^2 - x + 2}{x^2 + x + 2}\,dx = x - \log(f(x)) + \dfrac{2}{\sqrt{7}}\tan^{-1}(g(x)) + c$, then $f(-1) + \sqrt{7}\,g(-1) =$ (A) $1$ (B) $0$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Splitting the numerator to isolate the derivative of the denominator identifies $f(x)=x^2+x+2$ and $g(x)=\frac{2x+1}{\sqrt7}$, giving $f(-1)+\sqrt7 g(-1)=1$.

   Concept and Intuition

   For a rational function whose numerator and denominator are both quadratics with the same leading coefficient, subtracting the denominator from the numerator strips off the constant term, leaving a simpler linear-over-quadratic integral that splits into a logarithm part (from the derivative of the denominator) and an arctan part (from completing the square).

   Step-by-Step Solution

   1. Note $x^2+x+2 - 2x = x^2-x+2$, so $\dfrac{x^2-x+2}{x^2+x+2} = 1 - \dfrac{2x}{x^2+x+2}$.
   2. $\displaystyle\int 1\,dx = x$.
   3. For $\displaystyle\int\frac{2x}{x^2+x+2}dx$, write $2x = (2x+1)-1$:

   • $\displaystyle\int\frac{2x+1}{x^2+x+2}dx = \log(x^2+x+2)$ (numerator is exactly the derivative of the denominator).
   • $\displaystyle\int\frac{-1}{x^2+x+2}dx = -\int\frac{dx}{(x+\frac12)^2+\frac74} = -\frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{2x+1}{\sqrt7}\right)$.

   4. So $\displaystyle\int\frac{2x}{x^2+x+2}dx = \log(x^2+x+2) - \frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{2x+1}{\sqrt7}\right)$. …

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9. 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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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin x - \cos x}{\sqrt{\sin 2x}}\,dx =$ (A) $-\log|\sin x - \cos x + \sqrt{\sin 2x}| + c$ (B) $-\log|\sin x + \cos x - \sqrt{\sin 2x}| + c$ (C) $-\log|\sin x + \cos x + \sqrt{\sin 2x}| + c$ (D) $-\log|\sin x - \cos x - \sqrt{\sin 2x}| + c$

    ›Reveal solutionSolution

    The key substitution is $t=\sin x+\cos x$, which turns $\sin 2x$ into $t^2-1$ and the numerator into $-dt$, reducing the integral to a standard $\int \dfrac{dt}{\sqrt{t^2-1}}$ form.

    Concept and Intuition

    Expressions like $\sin x \pm \cos x$ paired with $\sin 2x$ are a classic signal to substitute $t=\sin x\pm\cos x$, because $(\sin x+\cos x)^2 = 1+\sin2x$ and $(\sin x-\cos x)^2=1-\sin2x$ — this converts everything to a single variable.

    Step-by-Step Solution

    1. Let $t=\sin x+\cos x$. Then $dt = (\cos x-\sin x)\,dx = -(\sin x-\cos x)\,dx$.
    2. Also $t^2 = \sin^2x+\cos^2x+2\sin x\cos x = 1+\sin2x$, so $\sin2x = t^2-1$, i.e. $\sqrt{\sin2x}=\sqrt{t^2-1}$.
    3. Rewrite the integral:

    $\int \frac{\sin x-\cos x}{\sqrt{\sin2x}}\,dx = \int \frac{-dt}{\sqrt{t^2-1}} = -\log\left|t+\sqrt{t^2-1}\right|+c.$

    4. Substitute back $t=\sin x+\cos x$ and $\sqrt{t^2-1}=\sqrt{\sin2x}$: …

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11. 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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12. 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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