Q.$\int \sqrt{1+x^2} dx$ is equal to (A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log |x+\sqrt{1+x^2}| + C$ (B) $\frac{2}{3}(1+x^2)^{\frac{3}{2}} + C$ (C) $\frac{2}{3}x(1+x^2)^{\frac{3}{2}} + C$ (D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log |x+\sqrt{1+x^2}| + C$

Showing the 12 most recent of 35 on this concept.

1. 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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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 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\, dx =$ (A) $\log|x| - \dfrac{1}{4x^4} + C$ (B) $\log|x| + \dfrac{1}{4x^4} + C$ (C) $\log|x| - \dfrac{4}{x^4} + C$ (D) $\log|x| + \dfrac{4}{x^4} + C$

    ›Reveal solutionSolution

    The expression under the square root is a perfect square, simplifying the integral to a sum of two elementary power terms.

    Concept and Intuition

    Recognize $x^4+x^{-4}+2$ as $(x^2+x^{-2})^2$ using the identity $a^2+b^2+2ab=(a+b)^2$ with $a=x^2,b=x^{-2}$.

    Step-by-Step Solution

    1. $(x^2+x^{-2})^2 = x^4+2+x^{-4}$ — matches the expression under the square root exactly.
    2. $\sqrt{x^4+x^{-4}+2} = |x^2+x^{-2}| = x^2+\dfrac{1}{x^2}$ (always positive for real $x\ne0$).
    3. Divide by $x^3$: $\dfrac{x^2+1/x^2}{x^3} = \dfrac1x+\dfrac{1}{x^5}$. …

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12. 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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