Q.Evaluate the definite integral: $\int_2^3 \frac{1}{x} \ dx$

Showing the 12 most recent of 29 on this concept.

1. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.$\displaystyle\int_{e}^{e^2} \dfrac{dx}{x \log x} =$ (A) $-1 + \log 2$ (B) $-2 + \log 3$ (C) $-\log 3$ (D) $\log 2$

   ›Reveal solutionSolution

   With $u=\log x$, $\displaystyle\int_e^{e^2}\frac{dx}{x\log x}=\int_1^2\frac{du}{u}=\log 2$.

   Concept and Intuition

   Whenever you see $\dfrac{dx}{x}$ alongside $\log x$, the substitution $u=\log x$ (so $du=dx/x$) is immediate — it converts the whole integral into $\int du/u$.

   Step-by-Step Solution

   1. Let $u=\log x\Rightarrow du=\dfrac{dx}{x}$.
   2. Limits: when $x=e$, $u=1$; when $x=e^2$, $u=2$.
   3. Integral becomes $\displaystyle\int_1^2\frac{du}{u}=[\log u]_1^2=\log 2-\log 1=\log 2$.

   Common Mistakes

   • Forgetting to change the limits of integration to the new variable $u$.
   • Leaving an extra factor of $x$ unaccounted for.

   ✓Final answer

   The correct option is (D) — $\log 2$.

   ANSWER: D

2. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.Let $f(x)=\displaystyle\int \dfrac{x}{(x^2+1)(x^2+3)}\,dx$. If $f(3)=\dfrac{1}{4}\log\left(\dfrac{5}{6}\right)$ then $f(0)=$ (A) $\dfrac{1}{4}\log\left(\dfrac{1}{3}\right)$ (B) $0$ (C) $\dfrac{1}{2}\log\left(\dfrac{1}{3}\right)$ (D) $\log\left(\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   Evaluating the integral in closed form and pinning the constant using $f(3)$ shows $C=0$, so $f(0)=\dfrac14\log\left(\dfrac13\right)$.

   Concept and Intuition

   Once we find the general antiderivative (up to an unknown constant $C$), any single given value of the function — here $f(3)$ — lets us solve for $C$ exactly, after which $f$ is completely determined and we can evaluate it anywhere, including at $x=0$.

   Step-by-Step Solution

   1. Substitute $u=x^2$, $du=2x\,dx$: $\displaystyle\int\dfrac{x\,dx}{(x^2+1)(x^2+3)} = \dfrac12\int\dfrac{du}{(u+1)(u+3)}$.
   2. Partial fractions: $\dfrac{1}{(u+1)(u+3)}=\dfrac{1/2}{u+1}-\dfrac{1/2}{u+3}$.
   3. So $\dfrac12\int\left[\dfrac{1/2}{u+1}-\dfrac{1/2}{u+3}\right]du = \dfrac14\left[\log(u+1)-\log(u+3)\right]+C = \dfrac14\log\left(\dfrac{u+1}{u+3}\right)+C$.
   4. Back-substitute $u=x^2$: $f(x)=\dfrac14\log\left(\dfrac{x^2+1}{x^2+3}\right)+C$.
   5. At $x=3$: $f(3)=\dfrac14\log\left(\dfrac{10}{12}\right)+C=\dfrac14\log\left(\dfrac56\right)+C$. Given $f(3)=\dfrac14\log\left(\dfrac56\right)$, we get $C=0$.
   6. So $f(0)=\dfrac14\log\left(\dfrac{0+1}{0+3}\right)=\dfrac14\log\left(\dfrac13\right)$.

   Common Mistakes

   • Forgetting the constant of integration entirely and assuming the "bare" antiderivative automatically satisfies $f(3)$.
   • Arithmetic slip simplifying $10/12$ to $5/6$.

   ✓Final answer

   The correct option is (A) — $\dfrac14\log\left(\dfrac13\right)$.

   ANSWER: A

3. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.$\displaystyle\int_{0}^{1} \dfrac{2x + 5}{x^2 + 3x + 2}\, dx =$ (A) $\log\left(\dfrac{16}{3}\right)$ (B) $0$ (C) $\log\left(\dfrac{3}{16}\right)$ (D) $4\log 2 - 2\log 3$

   ›Reveal solutionSolution

   Partial fractions turn $\frac{2x+5}{x^2+3x+2}$ into $\frac{3}{x+1}-\frac{1}{x+2}$, and evaluating from $0$ to $1$ gives $\log(16/3)$.

   Concept and Intuition

   A rational function whose denominator factors into distinct linear terms always splits into a sum of simple $\dfrac{\text{const}}{\text{linear}}$ pieces, each of which integrates to a logarithm. This is the bread-and-butter technique for rational integrands.

   Step-by-Step Solution

   1. Factor the denominator: $x^2+3x+2=(x+1)(x+2)$.
   2. Write $\dfrac{2x+5}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$, so $2x+5=A(x+2)+B(x+1)$.
   3. At $x=-1$: $3=A(1)\Rightarrow A=3$. At $x=-2$: $1=B(-1)\Rightarrow B=-1$.
   4. So the integrand is $\dfrac{3}{x+1}-\dfrac{1}{x+2}$, and

   $\int_0^1\left(\frac{3}{x+1}-\frac{1}{x+2}\right)dx=\Big[3\log|x+1|-\log|x+2|\Big]_0^1.$

   5. At $x=1$: $3\ln2-\ln3$. At $x=0$: $3\ln1-\ln2=-\ln2$.
   6. Difference: $(3\ln2-\ln3)-(-\ln2)=4\ln2-\ln3=\log(2^4)-\ln3=\log\dfrac{16}{3}$.

   Common Mistakes

   • Sign error when subtracting the lower limit (forgetting $-\ln2$ at $x=0$ flips to $+\ln2$).
   • Confusing this with $4\log2-2\log3=\log(16/9)$, a distractor option from a wrong partial-fraction split.

   ✓Final answer

   The correct option is (A) — $\log\left(\dfrac{16}{3}\right)$.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int_{\sqrt{2}}^{2} \dfrac{x}{(x^3 - x^2 + x - 1)(x + 1)}\, dx =$ (A) $\dfrac{1}{2}\log\left(\dfrac{9}{4}\right)$ (B) $\dfrac{1}{4}\log\left(\dfrac{9}{5}\right)$ (C) $2\log 3$ (D) $3\log 2$

   ›Reveal solutionSolution

   The cubic factors nicely, and the whole denominator collapses to $x^4-1$, turning the integral into a standard $u=x^2$ substitution followed by a log-of-ratio evaluation. Answer: $\dfrac14\log\left(\dfrac95\right)$.

   Concept and Intuition

   Cubic factorisation by grouping ($x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1)$) is the key simplification; once multiplied by the extra $(x+1)$ factor, the denominator becomes the recognisable difference-of-squares product $(x^2-1)(x^2+1)=x^4-1$, and $\int x/(x^4-1)\,dx$ is a textbook $u=x^2$ substitution leading to a partial-fraction log form.

   Step-by-Step Solution

   1. Factor by grouping: $x^3-x^2+x-1=x^2(x-1)+1(x-1)=(x-1)(x^2+1)$.
   2. Denominator $=(x-1)(x^2+1)(x+1)=(x-1)(x+1)(x^2+1)=(x^2-1)(x^2+1)=x^4-1$.
   3. Integral $=\displaystyle\int_{\sqrt2}^{2}\frac{x}{x^4-1}dx$. Let $u=x^2$, $du=2x\,dx$; limits $x=\sqrt2\Rightarrow u=2$, $x=2\Rightarrow u=4$.
   4. $=\dfrac12\displaystyle\int_2^4\frac{du}{u^2-1}=\dfrac12\cdot\dfrac12\log\left|\dfrac{u-1}{u+1}\right|\Big|_2^4=\dfrac14\left[\log\dfrac{3}{5}-\log\dfrac13\right]$.
   5. $=\dfrac14\log\left(\dfrac{3/5}{1/3}\right)=\dfrac14\log\left(\dfrac95\right)$.

   Common Mistakes

   • Failing to spot the cubic factorisation and instead attempting a messy direct partial-fraction decomposition of a quartic-degree denominator.
   • Sign/limit slip when substituting $u=x^2$ and forgetting to convert the limits.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{4}\log\left(\dfrac{9}{5}\right)$.

   ANSWER: B

5. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x(x^5 + 3)} =$ (A) $\dfrac{1}{3} \log|x^5 + 3| + c$ (B) $\dfrac{1}{15} \log\left|\dfrac{x^5}{x^5 + 3}\right| + c$ (C) $\dfrac{1}{5} \log\left|\dfrac{x^5}{x^5 + 3}\right| + c$ (D) $\dfrac{1}{5} \log|x^5 + 3| + c$

   ›Reveal solutionSolution

   $\int\dfrac{dx}{x(x^n+a)}=\dfrac{1}{na}\log\left|\dfrac{x^n}{x^n+a}\right|+c$; here $n=5,\ a=3$.

   Concept and Intuition

   Multiplying by $x^{n-1}/x^{n-1}$ converts $\int \dfrac{dx}{x(x^n+a)}$ into a function of $t=x^n$ alone, reducing it to the standard $\int \dfrac{dt}{t(t+a)}$ partial-fraction integral.

   Step-by-Step Solution

   1. Multiply numerator and denominator by $x^4$: $\dfrac{1}{x(x^5+3)}=\dfrac{x^4}{x^5(x^5+3)}$.
   2. Let $t=x^5\Rightarrow dt=5x^4\,dx$, so $x^4\,dx=\dfrac{dt}{5}$.
   3. Integral becomes $\dfrac15\displaystyle\int\frac{dt}{t(t+3)}$.
   4. Partial fractions: $\dfrac{1}{t(t+3)}=\dfrac13\left(\dfrac1t-\dfrac1{t+3}\right)$.
   5. So integral $=\dfrac15\cdot\dfrac13\left[\log|t|-\log|t+3|\right]+c=\dfrac1{15}\log\left|\dfrac{t}{t+3}\right|+c$.
   6. Substitute back $t=x^5$: $\dfrac{1}{15}\log\left|\dfrac{x^5}{x^5+3}\right|+c$.

   Common Mistakes

   • Using the wrong power of $x$ to multiply through (must be $x^{n-1}=x^4$, not $x$).
   • Missing the $\tfrac15$ factor from $dt=5x^4dx$.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{15}\log\left|\dfrac{x^5}{x^5+3}\right|+c$.

   ANSWER: B

6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\cos x}\left[\dfrac{1}{\sin x}-\dfrac{1}{\sin x+3\cos x}\right]dx=$ (A) $\dfrac{1}{3}\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$ (B) $\log\left|\dfrac{\cos x}{\sin x+3\cos x}\right|+c$ (C) $\dfrac{1}{3}\log\left|\dfrac{\cos x}{\sin x+3\cos x}\right|+c$ (D) $\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$

   ›Reveal solutionSolution

   Simplifying the bracket first collapses the integrand to a single rational trig fraction, which becomes an easy partial-fraction integral in $\tan x$.

   Concept and Intuition

   Combining the two fractions inside the bracket over a common denominator cancels the $\sin x$ term neatly, leaving a $\cos x$ in the numerator that cancels the outer $1/\cos x$ — a strong hint to simplify algebraically before attempting any substitution.

   Step-by-Step Solution

   1. $\dfrac{1}{\sin x}-\dfrac{1}{\sin x+3\cos x}=\dfrac{(\sin x+3\cos x)-\sin x}{\sin x(\sin x+3\cos x)}=\dfrac{3\cos x}{\sin x(\sin x+3\cos x)}$.
   2. Multiplying by the outer $\dfrac{1}{\cos x}$: integrand $=\dfrac{3}{\sin x(\sin x+3\cos x)}$.
   3. Divide numerator and denominator by $\cos^2x$: $=\dfrac{3\sec^2x}{\tan x(\tan x+3)}$.
   4. Let $u=\tan x$, $du=\sec^2x\,dx$: integral $=\displaystyle\int\frac{3}{u(u+3)}du$.
   5. Partial fractions: $\dfrac{3}{u(u+3)}=\dfrac1u-\dfrac{1}{u+3}$, so the integral is $\log|u|-\log|u+3|+c=\log\left|\dfrac{u}{u+3}\right|+c$.
   6. Substituting back $u=\tan x$: $\log\left|\dfrac{\tan x}{\tan x+3}\right|+c=\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$.

   Common Mistakes

   • Attempting substitution before simplifying the bracket, leading to an unnecessarily messy integral.
   • Losing track of the factor from partial fractions and introducing a spurious $\frac13$.

   ✓Final answer

   The correct option is (D) — $\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{1/2}^{1/\sqrt{2}} \dfrac{1}{(x+\sqrt{1-x^2})(1-x^2)}\,dx =$ (A) $\log(\sqrt{3}+1)$ (B) $\log(\sqrt{3}-1)$ (C) $\log(3+\sqrt{3})$ (D) $\log(3-\sqrt{3})$

   ›Reveal solutionSolution

   The substitution $x=\sin\theta$ turns the awkward algebraic integrand into $\sec^2\theta/(1+\tan\theta)$, a direct log integral. Answer: $\log(3-\sqrt3)$.

   Concept and Intuition

   Whenever $\sqrt{1-x^2}$ appears together with $x$ in a rational expression, the trigonometric substitution $x=\sin\theta$ is the natural move — it turns $\sqrt{1-x^2}$ into $\cos\theta$ and $dx$ into $\cos\theta\,d\theta$, often producing large cancellations.

   Step-by-Step Solution

   1. Let $x=\sin\theta$, $dx=\cos\theta\,d\theta$, $\sqrt{1-x^2}=\cos\theta$, $1-x^2=\cos^2\theta$.
   2. Limits: $x=\tfrac12\Rightarrow\theta=\tfrac\pi6$; $x=\tfrac1{\sqrt2}\Rightarrow\theta=\tfrac\pi4$.
   3. The integral becomes

   $\int \frac{\cos\theta\,d\theta}{(\sin\theta+\cos\theta)\cos^2\theta} = \int\frac{d\theta}{\cos\theta(\sin\theta+\cos\theta)} = \int\frac{d\theta}{\sin\theta\cos\theta+\cos^2\theta}.$

   4. Divide numerator and denominator by $\cos^2\theta$: $\displaystyle\int \frac{\sec^2\theta\,d\theta}{\tan\theta+1}$.
   5. Let $t=\tan\theta$, $dt=\sec^2\theta\,d\theta$: $\displaystyle\int\frac{dt}{t+1} = \log|t+1|$.
   6. Evaluate from $\theta=\pi/6$ ($t=1/\sqrt3$) to $\theta=\pi/4$ ($t=1$):

   $\log(1+1) - \log\!\left(1+\frac{1}{\sqrt3}\right) = \log 2 - \log\!\left(\frac{\sqrt3+1}{\sqrt3}\right) = \ln2 + \log\sqrt3 - \log(\sqrt3+1) = \log\!\left(\frac{2\sqrt3}{\sqrt3+1}\right).$

   7. Rationalize: $\dfrac{2\sqrt3}{\sqrt3+1}\cdot\dfrac{\sqrt3-1}{\sqrt3-1} = \dfrac{2\sqrt3(\sqrt3-1)}{2} = \sqrt3(\sqrt3-1) = 3-\sqrt3$.
   8. So the value is $\log(3-\sqrt3)$.

   Common Mistakes

   • Forgetting to rationalize the final expression, leaving it in a form that doesn't visibly match any option.
   • Losing track of the limits after the substitution $t=\tan\theta$ (mixing up $\theta$-limits with $t$-limits).

   ✓Final answer

   The correct option is (D) — $\log(3-\sqrt{3})$.

   ANSWER: D

8. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $f_n(x) = \displaystyle\int \dfrac{1-\log x}{x^n}\,dx$, then $f_2(e) - f_3(e) =$ (A) $\dfrac{1}{4e^2}[4e-1] + c$ (B) $\dfrac{1}{e^2} - \dfrac{2}{e} + c$ (C) $\dfrac{e-1}{e+1} + c$ (D) $\dfrac{1}{2e}\left[\dfrac{1+e}{e}\right] + c$

   ›Reveal solutionSolution

   This tests integration by parts on a family $f_n(x)=\int(1-\log x)x^{-n}dx$ and simplifying $f_2(e)-f_3(e)$; the answer is $\dfrac{4e-1}{4e^2}$.

   Concept and Intuition

   The integrand $(1-\log x)x^{-n}$ is exactly the derivative of $\dfrac{\log x}{x^{n-1}}$ type expressions once you notice the pattern: differentiating $\dfrac{\log x}{x^{k}}$ produces a $\dfrac{1}{x^{k+1}}$ term and a $-k\dfrac{\log x}{x^{k+1}}$ term, which combine nicely for specific $n$. Rather than guess, integrating by parts systematically gives a general formula valid for every $n\ne 1$.

   Step-by-Step Solution

   1. Write $f_n(x)=\int (1-\log x)x^{-n}dx$. Take $u=1-\log x$, $dv=x^{-n}dx$, so $du=-\dfrac1x dx$, $v=\dfrac{x^{1-n}}{1-n}$.
   2. By parts: $f_n(x)=\dfrac{(1-\log x)x^{1-n}}{1-n}-\int \dfrac{x^{1-n}}{1-n}\left(-\dfrac1x\right)dx=\dfrac{(1-\log x)x^{1-n}}{1-n}+\dfrac{1}{1-n}\int x^{-n}dx$.
   3. $\int x^{-n}dx=\dfrac{x^{1-n}}{1-n}$, so $f_n(x)=\dfrac{x^{1-n}}{1-n}\left[(1-\log x)+\dfrac{1}{1-n}\right]+C$.
   4. For $n=2$ ($1-n=-1$): $f_2(x)=-x^{-1}\left[(1-\log x)-1\right]=-x^{-1}(-\log x)=\dfrac{\log x}{x}+C$.
   5. For $n=3$ ($1-n=-2$): $f_3(x)=\dfrac{x^{-2}}{-2}\left[(1-\log x)-\dfrac12\right]=\dfrac{x^{-2}}{-2}\left(\dfrac12-\log x\right)=\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+C$.
   6. Evaluate at $x=e$: $f_2(e)=\dfrac{1}{e}$, $f_3(e)=\dfrac{1}{2e^2}-\dfrac{1}{4e^2}=\dfrac{1}{4e^2}$.
   7. $f_2(e)-f_3(e)=\dfrac1e-\dfrac{1}{4e^2}=\dfrac{4e-1}{4e^2}=\dfrac{1}{4e^2}[4e-1]$.

   Common Mistakes

   • Forgetting the extra $\dfrac{1}{1-n}$ term generated from integrating $x^{-n}$ again during parts.
   • Sign errors when $1-n$ is negative (as it is for both $n=2,3$).

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{4e^2}[4e-1]+c$.

   ANSWER: A

9. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{1+\cos x+\sin x}\,dx =$ (A) $\dfrac{\pi}{2}+\dfrac{1}{2}\log 2$ (B) $\dfrac{\pi}{4}-\dfrac{1}{2}\log 2$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{3\pi}{4}+\log 2$

   ›Reveal solutionSolution

   This tests the classic $I=J$ symmetry trick combined with the Weierstrass substitution; the answer is (B).

   Concept and Intuition

   When an integral is symmetric under $x\to \pi/2-x$ (swapping sine and cosine roles) but the numerator only has one of them, pairing it with its "partner" integral (numerator = the other function) lets you add the two to get something tractable, using $I=J$ from the symmetry.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\frac{\sin x}{1+\sin x+\cos x}dx$, $J=\displaystyle\int_0^{\pi/2}\frac{\cos x}{1+\sin x+\cos x}dx$.
   2. Substituting $x\to \pi/2-x$ in $I$ turns it into $J$, so $I=J$.
   3. $I+J=\displaystyle\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sin x+\cos x}dx=\int_0^{\pi/2}\left[1-\frac{1}{1+\sin x+\cos x}\right]dx=\frac{\pi}{2}-K$, where $K=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sin x+\cos x}$.
   4. Compute $K$ using the Weierstrass substitution $t=\tan(x/2)$: $\sin x=\frac{2t}{1+t^2}$, $\cos x=\frac{1-t^2}{1+t^2}$, $dx=\frac{2dt}{1+t^2}$. Then $1+\sin x+\cos x=\frac{2(1+t)}{1+t^2}$, so the integrand simplifies to $\frac{dt}{1+t}$.
   5. Limits: $x:0\to\pi/2$ gives $t:0\to1$. So $K=\int_0^1\frac{dt}{1+t}=\log 2$.
   6. Since $I=J$, $2I=\frac{\pi}{2}-\ln2$, so $I=\frac{\pi}{4}-\frac12\ln2$.

   Common Mistakes

   • Forgetting to check $I=J$ before using $2I=I+J$.
   • Errors in the Weierstrass substitution algebra (sign of $1-t^2$).

   ✓Final answer

   The correct option is (B) — $\dfrac{\pi}{4}-\dfrac{1}{2}\log 2$.

   ANSWER: B

10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{\cos x + \sqrt{3}\sin x}=$ (A) $\log\left(\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right)+c$ (B) $\log\left(\tan\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)\right)+c$ (C) $\dfrac{1}{2}\log\left(\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right)+c$ (D) $\dfrac{1}{2}\log\left(\tan\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)\right)+c$

    ›Reveal solutionSolution

    Rewriting $\cos x+\sqrt3\sin x$ as $2\sin(x+\pi/6)$ turns the integral into the standard $\int\csc\theta\,d\theta$ form, giving $\frac12\log\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)+c$.

    Concept and Intuition

    Any expression $a\cos x+b\sin x$ can be written as $R\sin(x+\phi)$ (or $R\cos(x-\phi)$) where $R=\sqrt{a^2+b^2}$. This is the key move that converts a linear combination of sine and cosine into the standard $\int\csc\theta\,d\theta = \log|\tan(\theta/2)|+c$ form.

    Step-by-Step Solution

    1. $\cos x + \sqrt3\sin x = 2\left(\dfrac12\cos x + \dfrac{\sqrt3}{2}\sin x\right) = 2\left(\sin x\cos\dfrac{\pi}{6}+\cos x\sin\dfrac{\pi}{6}\right) = 2\sin\left(x+\dfrac{\pi}{6}\right)$.
    2. So $\displaystyle\int\dfrac{dx}{\cos x+\sqrt3\sin x} = \dfrac12\int\csc\left(x+\dfrac{\pi}{6}\right)dx$.
    3. Using $\displaystyle\int\csc\theta\,d\theta = \log\left|\tan\dfrac{\theta}{2}\right|+c$ with $\theta = x+\dfrac{\pi}{6}$: $= \dfrac12\log\left|\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right|+c$.

    Common Mistakes

    • Using $R\cos(x-\phi)$ form instead and mismatching the phase shift, leading to $\pi/12$ turning into a different fraction.
    • Forgetting the leading factor of $\dfrac12$ that comes from $R=2$.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{2}\log\left(\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right)+c$.

    ANSWER: C

11. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{e^{2x}-1}{e^{2x}+e^x+1}\,dx =$ (A) $\log(e^{2x}+e^x+1) + x + c$ (B) $\log(e^{2x}+e^x+1) - x + c$ (C) $\log\left(\dfrac{e^{2x}+e^x+1}{e^{2x}}\right) + c$ (D) $\log\left|\dfrac{e^{2x}+e^x+1}{e^{2x}-1}\right| + c$

    ›Reveal solutionSolution

    This tests recognizing a disguised $\int \frac{f'(x)}{f(x)}dx$ form after dividing through by $e^x$; the answer is $\log(e^{2x}+e^x+1)-x+c$.

    Concept and Intuition

    Whenever an integrand is a ratio of exponentials, dividing numerator and denominator by the smallest power often reveals a $\dfrac{du}{u}$ structure, since exponential derivatives reproduce exponentials. Here $e^{2x}+e^x+1$ divided by $e^x$ becomes $e^x+1+e^{-x}$, and its derivative is $e^x-e^{-x}$ — exactly the (rescaled) numerator.

    Step-by-Step Solution

    1. Divide numerator and denominator of $\dfrac{e^{2x}-1}{e^{2x}+e^x+1}$ by $e^x$:

    $\frac{e^{2x}-1}{e^{2x}+e^x+1} = \frac{e^x - e^{-x}}{e^x+1+e^{-x}}$

    2. Let $u = e^x + e^{-x} + 1$. Then $\dfrac{du}{dx} = e^x - e^{-x}$, which is exactly the numerator.
    3. So the integral becomes $\displaystyle\int \frac{du}{u} = \log|u| + c = \log(e^x+e^{-x}+1) + c$.
    4. Rewrite in terms of $e^{2x}$: multiply inside the log by $\dfrac{e^x}{e^x}$:

    $e^x+e^{-x}+1 = \frac{e^{2x}+1+e^x}{e^x}$

    5. So $\log(e^x+e^{-x}+1) = \log(e^{2x}+e^x+1) - \log(e^x) = \log(e^{2x}+e^x+1) - x$.
    6. Final result: $\log(e^{2x}+e^x+1) - x + c$.
    7. Check by differentiating: $\dfrac{d}{dx}\big[\log(e^{2x}+e^x+1)-x\big] = \dfrac{2e^{2x}+e^x}{e^{2x}+e^x+1} - 1 = \dfrac{2e^{2x}+e^x-e^{2x}-e^x-1}{e^{2x}+e^x+1} = \dfrac{e^{2x}-1}{e^{2x}+e^x+1}$, which matches the original integrand exactly.

    Common Mistakes

    • Trying to integrate directly without dividing by $e^x$ first — the denominator's true derivative structure is hidden until this step.
    • Forgetting the $-x$ term that comes from converting $\log(e^x+e^{-x}+1)$ back into $\log(e^{2x}+e^x+1)$ form, which flips option (A) into the wrong sign.

    ✓Final answer

    The correct option is (B) — $\log(e^{2x}+e^x+1) - x + c$.

    ANSWER: B

12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.The parametric form of a curve is $x = \frac{t^3}{t^2-1}$, $y = \frac{t}{t^2-1}$, then $\int \frac{dx}{x - 3y} =$ (A) $\frac{1}{2}\log(t^2 - 1) + C$ (B) $2\log(t(t^2-1)) + C$ (C) $\frac{1}{4}\log\left(\frac{t}{t^2-3}\right) + C$ (D) $\frac{5}{2}\log\left(t + \frac{1}{t^2}\right) + C$

    ›Reveal solutionSolution

    A parametric-curve integral that collapses to a simple $\int \frac{t\,dt}{t^2-1}$ once $x-3y$ and $dx$ are both written in terms of $t$; answer is $\frac12\log(t^2-1)+C$.

    Concept and Intuition

    For a parametric curve, any integral $\int F(x,y)\,dx$ becomes $\int F(x(t),y(t))\,x'(t)\,dt$ — a pure function of $t$. Simplify $x-3y$ first since it often factors nicely with $x'(t)$.

    Step-by-Step Solution

    1. $x - 3y = \dfrac{t^3}{t^2-1} - \dfrac{3t}{t^2-1} = \dfrac{t^3-3t}{t^2-1} = \dfrac{t(t^2-3)}{t^2-1}$.
    2. Differentiate $x=\dfrac{t^3}{t^2-1}$ by the quotient rule:

    $\frac{dx}{dt}=\frac{3t^2(t^2-1)-t^3(2t)}{(t^2-1)^2}=\frac{3t^4-3t^2-2t^4}{(t^2-1)^2}=\frac{t^4-3t^2}{(t^2-1)^2}=\frac{t^2(t^2-3)}{(t^2-1)^2}.$

    3. So $\frac{dx}{x-3y}=\frac{\dfrac{t^2(t^2-3)}{(t^2-1)^2}}{\dfrac{t(t^2-3)}{t^2-1}}\,dt=\frac{t^2(t^2-3)}{(t^2-1)^2}\cdot\frac{t^2-1}{t(t^2-3)}\,dt=\frac{t}{t^2-1}\,dt.$
    4. $\displaystyle\int \frac{t}{t^2-1}\,dt = \frac12\log|t^2-1|+C$ (substitute $u=t^2-1$, $du=2t\,dt$).

    Common Mistakes

    • Forgetting to cancel the common factor $(t^2-3)$ between numerator and denominator, which makes the integral look far messier than it is.
    • Sign errors in the quotient-rule differentiation of $x(t)$.

    ✓Final answer

    The correct option is (A) — $\frac{1}{2}\log(t^2-1)+C$.

    ANSWER: A