Q.Evaluate: $\int_{1}^{2} \dfrac{dx}{\sqrt{(x-1)(2-x)}}$

Showing the 12 most recent of 35 on this concept.

1. 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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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{1}{x^2 + 4x + \alpha}\,dx = \dfrac{1}{2\sqrt{2}}\mathrm{Tan}^{-1}\left(\dfrac{x+2}{2\sqrt{2}}\right) + c$, then $\displaystyle\int \dfrac{1}{x^2 + 4x - \alpha}\,dx =$ (A) $\dfrac{1}{4}\log\left(\dfrac{x-12}{x+2}\right) + k$ (B) $\dfrac{1}{8}\log\left(\dfrac{x-2}{x+6}\right) + k$ (C) $\dfrac{1}{8}\log\left(\dfrac{x+6}{x+8}\right) + k$ (D) $\dfrac{1}{4}\log\left(\dfrac{x-12}{x+16}\right) + k$

   ›Reveal solutionSolution

   First recover $\alpha=12$ by matching the given arctan antiderivative to the standard

   form; then integrate $\int\frac{dx}{x^2+4x-12}$ by partial fractions to get

   $\frac18\log\frac{x-2}{x+6}+k$.

   Concept and Intuition

   $\int\frac{dx}{u^2+k^2}=\frac1k\tan^{-1}(u/k)+c$ when the quadratic has no real roots

   (discriminant negative); $\int\frac{dx}{u^2-k^2}=\frac1{2k}\log\left|\frac{u-k}{u+k}\right|+c$

   when it factors into real linear pieces. The sign of the completed-square constant is

   what decides which family applies — here that same constant, $\alpha$, is first

   recovered from the arctan form, then reused with the opposite sign in the second

   integral.

   Step-by-Step Solution

   1. Complete the square: $x^2+4x+\alpha=(x+2)^2+(\alpha-4)$.
   2. Given antiderivative uses $\tan^{-1}\!\left(\frac{x+2}{2\sqrt2}\right)$ scaled by $\frac1{2\sqrt2}$ — matching $\frac1k\tan^{-1}(u/k)$ with $k=2\sqrt2$ requires $\alpha-4=k^2=(2\sqrt2)^2=8$, so $\alpha=12$.
   3. Now find $\displaystyle\int\dfrac{dx}{x^2+4x-12}$. Factor: $x^2+4x-12=(x+6)(x-2)$ (since $6\times(-2)=-12$ and $6+(-2)=4$).
   4. Partial fractions: $\dfrac{1}{(x+6)(x-2)}=\dfrac{A}{x-2}+\dfrac{B}{x+6}$. Solving, …

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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 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\tan x}{1 + \tan x + \tan^2 x}\, dx = x - \dfrac{K}{\sqrt{A}} \mathrm{Tan}^{-1}\left(\dfrac{K\tan x + 1}{\sqrt{A}}\right) + c$, then the ordered pair $(K, A) =$ (A) $(2, 3)$ (B) $(2, 1)$ (C) $(-2, 1)$ (D) $(-2, 3)$

   ›Reveal solutionSolution

   Rewrite the integrand using $\tan x=(1+\tan x+\tan^2x)-\sec^2x$ and substitute $t=\tan x$; the constants come out as $(K,A)=(2,3)$.

   Concept and Intuition

   When a rational function of $\tan x$ has $\sec^2x$ hiding in the numerator's structure, splitting the numerator to expose $1+\tan^2x=\sec^2x$ turns the whole thing into a substitution-ready form with $t=\tan x,\,dt=\sec^2x\,dx$.

   Step-by-Step Solution

   1. Write

   $\frac{\tan x}{1+\tan x+\tan^2x}=\frac{(1+\tan x+\tan^2x)-(1+\tan^2x)}{1+\tan x+\tan^2x}=1-\frac{\sec^2x}{1+\tan x+\tan^2x}.$

   2. So the integral is

   $\int 1\,dx-\int\frac{\sec^2x}{1+\tan x+\tan^2x}\,dx=x-\int\frac{dt}{1+t+t^2}\quad(t=\tan x).$

   3. Complete the square: $t^2+t+1=\left(t+\tfrac12\right)^2+\tfrac34$. So

   $\int\frac{dt}{(t+\frac12)^2+(\frac{\sqrt3}{2})^2}=\frac{1}{\sqrt3/2}\mathrm{Tan}^{-1}\left(\frac{t+\frac12}{\sqrt3/2}\right)=\frac{2}{\sqrt3}\mathrm{Tan}^{-1}\left(\frac{2t+1}{\sqrt3}\right).$

   4. Hence …

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5. 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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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 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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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{1}{9\cos^2x - 24\sin x\cos x + 16\sin^2 x}dx =$ (A) $\frac{\cos x}{4(3\cos x - 4\sin x)} + c$ (B) $\frac{\sin x}{4(3\cos x - 4\sin x)} + c$ (C) $\frac{\cos x}{3\cos x - 4\sin x} + c$ (D) $\frac{\sin x}{3\cos x - 4\sin x} + c$

   ›Reveal solutionSolution

   Recognising the denominator as the perfect square $(3\cos x-4\sin x)^2$ and substituting $u=3-4\tan x$ reduces the integral to a simple power rule, giving $\frac{\cos x}{4(3\cos x-4\sin x)}+c$.

   Concept and Intuition

   When a trigonometric denominator has the form $a^2\cos^2x - 2ab\sin x\cos x + b^2\sin^2x$, check whether it is a perfect square $(a\cos x - b\sin x)^2$ first — this instantly turns a messy-looking integral into $\int\sec^2(\cdot)/(\text{linear in }\tan x)^2\,dx$, solvable by substitution.

   Step-by-Step Solution

   1. Observe $9\cos^2x-24\sin x\cos x+16\sin^2x=(3\cos x-4\sin x)^2$ (matches $a^2-2ab+b^2$ with $a=3\cos x,\ b=4\sin x$).
   2. So the integral is $\displaystyle\int\frac{dx}{(3\cos x-4\sin x)^2}$.
   3. Divide numerator and denominator by $\cos^2x$: $=\displaystyle\int\frac{\sec^2x\,dx}{(3-4\tan x)^2}$.
   4. Let $u=3-4\tan x$, so $du=-4\sec^2x\,dx$, i.e. $\sec^2x\,dx=-\tfrac14du$. …

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

    ›Reveal solutionSolution

    The integral $\int(\sqrt{\tan x}+\sqrt{\cot x})dx$ is a classic "sum of square-root trig" integral that reduces to an arctangent form; verified here by direct differentiation, giving option (C).

    Concept and Intuition

    $\sqrt{\tan x}+\sqrt{\cot x}$ combines to $\dfrac{\tan x + 1}{\sqrt{\tan x}}$, and integrals of this shape are standard results that produce an inverse-tangent (or occasionally inverse-sine) antiderivative involving $\sqrt{2\tan x}$. Rather than re-deriving the substitution from scratch, the fastest reliable check with multiple-choice options is to differentiate each candidate and see which one reproduces the integrand exactly.

    Step-by-Step Solution

    1. Rewrite the integrand: $\sqrt{\tan x}+\sqrt{\cot x}=\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}=\dfrac{\tan x+1}{\sqrt{\tan x}}$.
    2. Test option (C): let $u=\dfrac{\tan x-1}{\sqrt{2\tan x}}$, and check $\dfrac{d}{dx}\left[\sqrt2\,\mathrm{Tan}^{-1}(u)\right]=\sqrt2\cdot\dfrac{u'}{1+u^2}$.
    3. Compute $1+u^2 = 1+\dfrac{(\tan x-1)^2}{2\tan x}=\dfrac{2\tan x+(\tan x-1)^2}{2\tan x}=\dfrac{\tan^2x+1}{2\tan x}=\dfrac{\sec^2x}{2\tan x}$.
    4. Compute $u'$ (quotient rule on $u=(\tan x-1)(2\tan x)^{-1/2}$):

    $u'=\sec^2x(2\tan x)^{-3/2}\big[(2\tan x)-( \tan x-1)\big] = \sec^2x(2\tan x)^{-3/2}(\tan x+1).$

    5. Combine: $\dfrac{u'}{1+u^2}=\sec^2x(2\tan x)^{-3/2}(\tan x+1)\cdot\dfrac{2\tan x}{\sec^2x}=\dfrac{2\tan x(\tan x+1)}{(2\tan x)^{3/2}}=\dfrac{\tan x+1}{\sqrt{2\tan x}}\cdot\dfrac{1}{1}$ (after simplifying $(2\tan x)^{3/2}/(2\tan x)=\sqrt{2\tan x}$), giving $\dfrac{\tan x+1}{\sqrt{2\tan x}}$. …

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11. 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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12. 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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