Q.Integrate the following function: $\sqrt{x^2+4x-5}$

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 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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3. 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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4. 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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5. 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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6. 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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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-2025-05-26-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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9. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If $I=\int_1^3\sqrt{3+x+x^2}\,dx$, then I lies in the interval (A) $(2\sqrt{5}, 2\sqrt{15})$ (B) $(\sqrt{3}, 2\sqrt{5})$ (C) $(\sqrt{23},\sqrt{33})$ (D) $(2\sqrt{15},\sqrt{23})$

   ›Reveal solutionSolution

   Bound the integral using the minimum and maximum of the increasing integrand on $[1,3]$: $I$ must lie between $2\sqrt5$ and $2\sqrt{15}$.

   Concept and Intuition

   When an exact antiderivative is messy, a quick way to pin down which interval an integral lies in is to bound the integrand: if $m\le f(x)\le M$ on $[a,b]$, then $m(b-a)\le\int_a^b f\,dx\le M(b-a)$, with strict inequality when $f$ is not constant.

   Step-by-Step Solution

   1. $x^2+x+3=(x+\tfrac12)^2+\tfrac{11}{4}$ is increasing for $x>-\tfrac12$, so on $[1,3]$, $f(x)=\sqrt{x^2+x+3}$ is strictly increasing.
   2. $f(1)=\sqrt{1+1+3}=\sqrt5$ (minimum on $[1,3]$).
   3. $f(3)=\sqrt{9+3+3}=\sqrt{15}$ (maximum on $[1,3]$).
   4. Since the interval length is $3-1=2$: $2\sqrt5 < I=\int_1^3 f(x)\,dx < 2\sqrt{15}$. …

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10. 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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11. 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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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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