Q.Integrate the following function: $\sqrt{x^2+4x+6}$

Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\displaystyle\int \left(\sqrt{x + \sqrt{12x - 36}} + \sqrt{x - \sqrt{12x - 36}}\right) dx =$ (A) $2\sqrt{3}x + C, \ \forall x$ (B) $\dfrac{4(x - 3)^{3/2}}{3} + C, \ \forall x$ (C) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & x > 6 \\ 2\sqrt{3}x + C, & 3 \leq x \leq 6 \end{cases}$ (D) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & 3 \leq x \leq 6 \\ 2\sqrt{3}x + C, & x > 6 \end{cases}$

   ›Reveal solutionSolution

   Substituting $t=\sqrt{x-3}$ turns both nested radicals into perfect squares $(t\pm\sqrt3)^2$; the integrand collapses to a constant $2\sqrt3$ on $[3,6]$ and to $2\sqrt{x-3}$ for $x>6$, giving the piecewise antiderivative in (C).

   Concept and Intuition

   Nested radicals of the form $\sqrt{x\pm\sqrt{\text{linear in }x}}$ often simplify to perfect squares under a substitution that removes the inner square root. Here the key is recognizing $12x-36$ is a perfect multiple of $(x-3)$.

   Step-by-Step Solution

   1. Let $t=\sqrt{x-3}\ge0$ (valid for $x\ge3$), so $x=t^2+3$ and $12x-36=12t^2\Rightarrow \sqrt{12x-36}=2\sqrt3\,t$.
   2. $x+\sqrt{12x-36} = t^2+3+2\sqrt3\,t = (t+\sqrt3)^2 \Rightarrow \sqrt{x+\sqrt{12x-36}} = t+\sqrt3$ (always non-negative).
   3. $x-\sqrt{12x-36} = t^2+3-2\sqrt3\,t=(t-\sqrt3)^2 \Rightarrow \sqrt{x-\sqrt{12x-36}} = |t-\sqrt3|$.
   4. Sum $= (t+\sqrt3)+|t-\sqrt3|$. If $t\ge\sqrt3$ (i.e. $x-3\ge3 \Rightarrow x\ge6$): sum $=2t=2\sqrt{x-3}$. If $0\le t<\sqrt3$ (i.e. $3\le x<6$): sum $=(t+\sqrt3)+(\sqrt3-t)=2\sqrt3$.
   5. Integrate each branch: $\int 2\sqrt{x-3}\,dx = \dfrac43(x-3)^{3/2}+C$ for $x\ge6$; $\int2\sqrt3\,dx = 2\sqrt3\,x+C$ for $3\le x\le6$.

   Common Mistakes

   • Dropping the absolute value on $\sqrt{(t-\sqrt3)^2}$ and getting a single formula valid everywhere (options (A),(B)) — the sign of $t-\sqrt3$ genuinely flips at $x=6$.
   • Mixing up which branch applies where.

   ✓Final answer

   The correct option is (C) — the piecewise antiderivative with $\dfrac43(x-3)^{3/2}+C$ for $x>6$ and $2\sqrt3x+C$ for $3\le x\le6$.

   ANSWER: C

2. 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 $u=x+3,\ a=4$:
   4. Result $=\sin^{-1}\left(\dfrac{x+3}{4}\right)+c$.

   Common Mistakes

   • Getting the sign of the completed square wrong (it must come out as $a^2-(\cdot)^2$, positive, for arcsine to apply — check by testing the vertex value).

   ✓Final answer

   The correct option is (C) — $\mathrm{Sin}^{-1}\left(\frac{x+3}{4}\right)+c$.

   ANSWER: C

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 $\dfrac ua = \dfrac{x+\frac12}{\sqrt3/2}=\dfrac{2x+1}{\sqrt3}$.
   5. So the integral $=\dfrac{2x+1}4\sqrt{x^2+x+1}+\dfrac38\mathrm{Sinh}^{-1}\!\left(\dfrac{2x+1}{\sqrt3}\right)+c$.

   Common Mistakes

   • Writing $\dfrac{x+1}4$ instead of $\dfrac{2x+1}4$ for the $u/2$ term (dropping the factor of 2 from $u=x+\tfrac12$).
   • Sign error on the $\mathrm{Sinh}^{-1}$ term.

   ✓Final answer

   The correct option is (A) — $\dfrac{2x+1}{4}\sqrt{x^2+x+1}+\dfrac38\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c$.

   ANSWER: A

4. 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 $x^3$: $\dfrac{x^2+1/x^2}{x^3} = \dfrac1x+\dfrac{1}{x^5}$.
   4. Integrate: $\int\left(\dfrac1x+x^{-5}\right)dx = \log|x| + \dfrac{x^{-4}}{-4}+C = \log|x|-\dfrac{1}{4x^4}+C$.

   Common Mistakes

   • Forgetting the absolute value / sign when taking the square root of a perfect square.

   ✓Final answer

   The correct option is (A) — $\log|x| - \dfrac{1}{4x^4} + C$.

   ANSWER: A

5. 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 $\dfrac12\sinh^{-1}\left(\dfrac{u}{a}\right)=\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)$.
   5. Combine: $\sqrt{x^2+x+1}+\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c$.

   Common Mistakes

   • Forgetting the factor of $\dfrac12$ in the split $x+1=\dfrac12(2x+1)+\dfrac12$.
   • Mixing up $\sinh^{-1}$ and $\cosh^{-1}$ — here the radicand $x^2+x+1$ is always positive (never zero), which is the $\sinh^{-1}$ (i.e. $x^2+a^2$) case, not the $\cosh^{-1}$ (i.e. $x^2-a^2$) case.

   ✓Final answer

   The correct option is (D) — $\sqrt{x^2+x+1}+\dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$.

   ANSWER: D

6. 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 $t=\sqrt{x-2}$:

   $\sqrt{x-2}-2\,\mathrm{Tan}^{-1}\!\left(\frac{\sqrt{x-2}}{2}\right)+c.$

   Common Mistakes

   • Forgetting the factor of $\dfrac12$ that comes from writing $2x+4=2(x+2)$ before substituting.
   • Losing the coefficient $4$ (as $\dfrac{4}{t^2+4}$, giving the $2\,\mathrm{Tan}^{-1}$ term) — a common slip is writing just $\mathrm{Tan}^{-1}(t/2)$ without the factor of $2$ out front.

   ✓Final answer

   The correct option is (B) — $\sqrt{x-2} - 2\,\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$.

   ANSWER: B

7. 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, $A(x+6)+B(x-2)=1$; at $x=2$: $8A=1\Rightarrow A=\tfrac18$; at $x=-6$: $-8B=1 \Rightarrow B=-\tfrac18$.
   5. Integrate: $\dfrac18\log|x-2|-\dfrac18\log|x+6|+k=\dfrac18\log\left|\dfrac{x-2}{x+6}\right|+k$.

   Common Mistakes

   • Using $\alpha=8$ (the value of $k^2$ itself) instead of correctly solving $\alpha-4=8$.
   • Sign errors in the partial-fraction split, e.g. writing $\frac{x+6}{x-2}$ (inverted) instead of $\frac{x-2}{x+6}$.

   ✓Final answer

   The correct option is (B) — $\dfrac18\log\left(\dfrac{x-2}{x+6}\right)+k$.

   ANSWER: B

8. 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 $\int\frac{du}{\sqrt{u^2+a^2}}=\mathrm{Sinh}^{-1}(u/a)$).
   6. Combine and substitute back $u=x-1$: $\sqrt{(x-1)^2+4}+\mathrm{Sinh}^{-1}\!\left(\dfrac{x-1}{2}\right)+c = \sqrt{x^2-2x+5}+\mathrm{Sinh}^{-1}\!\left(\dfrac{x-1}{2}\right)+c$.

   Common Mistakes

   • Splitting the numerator incorrectly (e.g. using $u-1$ instead of $u+1$ after the substitution, or forgetting the "+1" entirely).
   • Confusing $\mathrm{Sinh}^{-1}$ with $\mathrm{Sin}^{-1}$ (only valid for $\sqrt{a^2-u^2}$, not $\sqrt{u^2+a^2}$).

   ✓Final answer

   The correct option is (A) — $\sqrt{x^2-2x+5} + \mathrm{Sinh}^{-1}\left(\dfrac{x-1}{2}\right) + c$.

   ANSWER: A

9. 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: $8x^5-4x^3-8x^5+8x^3-4x=4x^3-4x$.
   6. So $F'=\dfrac{4x^3-4x}{4x^4u}=\dfrac{x^3-x}{x^4u}=\dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}$ — exactly the given integrand.

   Common Mistakes

   • Trying a messy trigonometric or algebraic substitution instead of the quicker verify-by-differentiation approach when options are given.
   • Sign errors while expanding $-4x(2x^4-2x^2+1)$.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2x^2}\sqrt{2x^4-2x^2+1}+C$.

   ANSWER: B

10. 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: $(2x^2-1)x^2-N = (2x^4-x^2)-(1-2x^2+2x^4) = x^2-1$.
    6. So $f'(x)=\dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}$ — exactly the integrand.

    Common Mistakes

    • Missing that $1-2x^2+2x^4$ and $2x^4-2x^2+1$ are the same polynomial (just written in different term order), which makes options (B) and (D) look different from the integrand's radicand when they aren't.
    • Trying a messy trig/algebraic substitution instead of the quicker verify-by-differentiation approach.

    ✓Final answer

    The correct option is (D) — $\dfrac{(1 - 2x^2 + 2x^4)^{1/2}}{2x^2} + c$.

    ANSWER: D

11. 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

    • Sign error inside the argument, giving $2x-1$ instead of $2x+1$ (that would come from completing the square of $x^2-x+1$, a different quadratic).
    • Dropping the factor of 2 from the original integrand, landing on $\frac{2}{\sqrt3}\tan^{-1}(\ldots)$ instead of $\frac{4}{\sqrt3}$.

    ✓Final answer

    The correct option is (B) — $\dfrac{4}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x + 1}{\sqrt{3}}\right) + c$.

    ANSWER: B

12. 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: $\dfrac{d}{du}\left[\dfrac{-u}{a^2\sqrt{u^2-a^2}}\right]=\dfrac{1}{(u^2-a^2)^{3/2}}$ ✓.
    4. Substitute back $u=x+a$, and $u^2-a^2=2ax+x^2$: result $=\dfrac{-(x+a)}{a^2\sqrt{2ax+x^2}}+C$.

    Common Mistakes

    • Mis-signing the standard antiderivative (dropping the leading minus sign).
    • Forgetting to substitute back from $u$ to $x$ inside the square root as well as outside it.

    ✓Final answer

    The correct option is (D) — $\dfrac{-1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$.

    ANSWER: D