Q.Find an anti derivative (or integral) of the function $e^{2x}$ by the method of inspection

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.$\displaystyle\int e^{-2x}\left(\dfrac{1 - \sin 2x}{1 + \cos 2x}\right) dx =$ (A) $\dfrac{1}{2}e^{-2x}\tan x + c$ (B) $-\dfrac{1}{2}e^{-2x}\tan x + c$ (C) $\dfrac{1}{2}e^{-2x}\cot x + c$ (D) $-\dfrac{1}{2}e^{-2x}\cot x + c$

   ›Reveal solutionSolution

   After simplifying the trig fraction, the integrand fits the pattern $e^{kx}[f'(x)+kf(x)]$, whose antiderivative is $e^{kx}f(x)$.

   Concept and Intuition

   $\dfrac{1-\sin2x}{1+\cos2x}=\dfrac{1-2\sin x\cos x}{2\cos^2x}=\dfrac{1}{2\cos^2x}-\dfrac{2\sin x\cos x}{2\cos^2x}=\dfrac12\sec^2x-\tan x$. This makes the integrand $e^{-2x}\left(\frac12\sec^2x-\tan x\right)$, which matches the standard product-rule-reversal shortcut $\int e^{kx}[f'(x)+kf(x)]\,dx=e^{kx}f(x)+C$.

   Step-by-Step Solution

   1. Simplify: $1+\cos2x=2\cos^2x$, $1-\sin2x=1-2\sin x\cos x$.
   2. $\dfrac{1-\sin2x}{1+\cos2x}=\dfrac{1}{2\cos^2x}-\dfrac{\sin x}{\cos x}=\dfrac12\sec^2x-\tan x$.
   3. Integral becomes $\displaystyle\int e^{-2x}\left(\dfrac12\sec^2x-\tan x\right)dx$.
   4. Take $f(x)=\dfrac12\tan x$, so $f'(x)=\dfrac12\sec^2x$, and $k=-2$.
   5. Check: $f'(x)+kf(x)=\dfrac12\sec^2x-2\cdot\dfrac12\tan x=\dfrac12\sec^2x-\tan x$ ✓ matches the bracket.
   6. By the shortcut, $\displaystyle\int e^{kx}[f'(x)+kf(x)]\,dx=e^{kx}f(x)+C$, so the integral $=e^{-2x}\cdot\dfrac12\tan x+C$.

   Common Mistakes

   • Forgetting the simplification $1+\cos2x=2\cos^2x$ and trying to integrate directly.
   • Sign error assembling $f'(x)+kf(x)$ and missing which sign of $k$ matches.

   ✓Final answer

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

   ANSWER: A

2. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.$\int \frac{e^{\cot x}}{\sin^2 x}(2\log\csc x + \sin 2x) \, dx =$ (A) $-2e^{\cot x}\log(\csc^2 x) + C$ (B) $-2e^{\cot x}\log(\csc x) + C$ (C) $-2e^{\cot x}\log(\csc x + \sin x) + C$ (D) $-2e^{\cot x}\log(\csc x - \cot x) + C$

   ›Reveal solutionSolution

   This tests recognizing $\int e^{f(x)}[g(x)+g'(x)\cdot(\text{chain factor})]dx$ patterns; the answer is $-2e^{\cot x}\log(\csc x)+C$.

   Concept and Intuition

   Many integrals of the form $\int e^{f(x)}h(x)\,dx$ simplify beautifully when $h(x)$ is secretly $u(x) - u'(x)/f'(x)$-type combination that matches a product-rule derivative. Here the key building blocks are $u=\cot x$ (so $du=-\csc^2x\,dx$) and the fact that $\frac{d}{dx}\log(\csc x) = -\cot x$.

   Step-by-Step Solution

   1. Rewrite the integrand: $\dfrac{e^{\cot x}}{\sin^2x}(2\log\csc x+\sin2x) = e^{\cot x}\csc^2x\big(2\log\csc x+\sin2x\big)$.
   2. Note $\csc^2x\sin2x = \dfrac{2\sin x\cos x}{\sin^2x} = 2\cot x$. So the integrand becomes

   $2\csc^2x\,e^{\cot x}\log(\csc x) + 2\cot x\,e^{\cot x}.$

   3. Consider $F(x) = -2e^{\cot x}\log(\csc x)$. Differentiate using the product rule:

   $F'(x) = -2\Big[(-\csc^2x)e^{\cot x}\log(\csc x) + e^{\cot x}(-\cot x)\Big] = 2\csc^2x\,e^{\cot x}\log(\csc x) + 2\cot x\,e^{\cot x}.$

   4. This is exactly the integrand from step 2. So $F(x)$ is the antiderivative.

   Common Mistakes

   • Forgetting the negative sign that comes from differentiating $\log(\csc x)$ (it gives $-\cot x$, not $\cot x$).
   • Not simplifying $\csc^2x\sin2x$ to $2\cot x$ before recognizing the pattern.

   ✓Final answer

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

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{e^{\sin x}(\sin 2x - 8\cos x)}{2(\sin x-3)^2}\,dx =$ (A) $e^{\sin x}(\sin x-3)+c$ (B) $\dfrac{e^{\sin x}}{(\sin x-3)^2}+c$ (C) $e^{\sin x}(\sin x-3)^2+c$ (D) $\dfrac{e^{\sin x}}{\sin x-3}+c$

   ›Reveal solutionSolution

   A "$\int e^{f(x)}[g(x)+g'(x)]$"-style integral in disguise — substitute $t=\sin x$ and recognize the resulting rational-times-exponential as the derivative of $\dfrac{e^t}{t-3}$. Answer: $\dfrac{e^{\sin x}}{\sin x-3}+c$.

   Concept and Intuition

   The identity $\displaystyle\int e^{u}\left[h(u)+h'(u)\right]du = e^u h(u)+c$ is the key trick behind many "exponential times trig" integrals. After substituting $t=\sin x$, the goal is to spot the integrand as $e^t\left[\dfrac1{t-3}-\dfrac1{(t-3)^2}\right]$, which is precisely $\dfrac{d}{dt}\!\left[\dfrac{e^t}{t-3}\right]$ by the quotient rule.

   Step-by-Step Solution

   1. Expand $\sin2x=2\sin x\cos x$ in the numerator:

   $e^{\sin x}(\sin2x-8\cos x) = e^{\sin x}\cos x(2\sin x-8) = 2e^{\sin x}\cos x(\sin x-4).$

   2. The integral becomes

   $\int \frac{2e^{\sin x}\cos x(\sin x-4)}{2(\sin x-3)^2}\,dx = \int \frac{e^{\sin x}\cos x(\sin x-4)}{(\sin x-3)^2}\,dx.$

   3. Substitute $t=\sin x \Rightarrow dt=\cos x\,dx$:

   $\int e^t\,\frac{t-4}{(t-3)^2}\,dt.$

   4. Decompose $\dfrac{t-4}{(t-3)^2}$ via partial fractions: write $t-4=A(t-3)+B$. At $t=3$: $B=-1$. Matching the $t$-coefficient: $A=1$. So $\dfrac{t-4}{(t-3)^2}=\dfrac{1}{t-3}-\dfrac{1}{(t-3)^2}$.
   5. Recognize the quotient-rule derivative: $\dfrac{d}{dt}\left[\dfrac{e^t}{t-3}\right] = \dfrac{e^t(t-3)-e^t}{(t-3)^2} = e^t\left[\dfrac{1}{t-3}-\dfrac{1}{(t-3)^2}\right]$ — exactly the integrand.
   6. Hence $\displaystyle\int e^t\,\frac{t-4}{(t-3)^2}\,dt = \frac{e^t}{t-3}+c$.
   7. Back-substitute $t=\sin x$: the answer is $\dfrac{e^{\sin x}}{\sin x-3}+c$.

   Common Mistakes

   • Missing the $\sin2x=2\sin x\cos x$ rewrite, which is essential to expose the $t=\sin x$ substitution.
   • Sign slip in the partial-fraction split of $\dfrac{t-4}{(t-3)^2}$.

   ✓Final answer

   The correct option is (D) — $\dfrac{e^{\sin x}}{\sin x-3}+c$.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{x-1}{(x+1)^3}e^x dx =$ (A) $\frac{e^x}{(x+1)^2} + c$ (B) $\frac{-e^x}{(x+1)^2} + c$ (C) $\frac{2e^x}{(x+1)} + c$ (D) $\frac{-e^x}{(x+1)^4} + c$

   ›Reveal solutionSolution

   The integrand fits the classic $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$ pattern with $f(x)=1/(x+1)^2$, giving the answer $\dfrac{e^x}{(x+1)^2}+c$ directly, with no further integration needed.

   Concept and Intuition

   Whenever an integrand has the shape $e^x$ times a sum of a function and its derivative, the antiderivative is simply $e^x$ times that function — because $\dfrac{d}{dx}\left[e^xf(x)\right]=e^x f(x)+e^xf'(x)=e^x[f(x)+f'(x)]$. Recognizing this pattern converts a seemingly hard rational-times-exponential integral into pure algebra: just guess $f(x)$ and check.

   Step-by-Step Solution

   1. Guess $f(x)=\dfrac{1}{(x+1)^2}$, motivated by the $(x+1)^3$ in the denominator (one power lower after differentiating).
   2. Compute $f'(x) = \dfrac{d}{dx}(x+1)^{-2} = -2(x+1)^{-3} = \dfrac{-2}{(x+1)^3}$.
   3. Form $f(x)+f'(x) = \dfrac{1}{(x+1)^2}-\dfrac{2}{(x+1)^3} = \dfrac{(x+1)-2}{(x+1)^3} = \dfrac{x-1}{(x+1)^3}$.
   4. This exactly matches the given integrand $\dfrac{x-1}{(x+1)^3}$.
   5. So $\displaystyle\int e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right]dx = e^x\cdot\frac{1}{(x+1)^2}+c$.

   Common Mistakes

   • Trying integration by parts directly on $(x-1)e^x/(x+1)^3$ instead of spotting the $e^x[f+f']$ structure, leading to a much longer (and error-prone) computation.
   • Sign slip when computing $f'(x)$, flipping the final sign of the answer.

   ✓Final answer

   The correct option is (A) — $\frac{e^x}{(x+1)^2} + c$.

   ANSWER: A

5. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{5\tan(x)}{\tan(x) - 2}\,dx = x + a\log|\sin(x) - 2\cos(x)| + k$, then $a =$ ______ (A) $-1$ (B) $-2$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   Convert $\tan x/(\tan x-2)$ into $\sin x/(\sin x-2\cos x)$, then decompose the numerator into a multiple of the denominator plus a multiple of its derivative.

   Concept and Intuition

   Rational functions of $\sin x,\cos x$ of this "linear-over-linear" type are standard-form integrals: writing the numerator as $A\cdot(\text{denominator}) + B\cdot(\text{derivative of denominator})$ splits the integral into a simple $x$-term plus a logarithm term.

   Step-by-Step Solution

   1. $\dfrac{5\tan x}{\tan x-2}=\dfrac{5\sin x/\cos x}{(\sin x-2\cos x)/\cos x}=\dfrac{5\sin x}{\sin x-2\cos x}$.
   2. Let denominator $D=\sin x-2\cos x$; $D'=\cos x+2\sin x$.
   3. Write $5\sin x=A\cdot D+B\cdot D'=A(\sin x-2\cos x)+B(\cos x+2\sin x)$.
   4. Matching $\sin x$: $5=A+2B$. Matching $\cos x$: $0=-2A+B\Rightarrow B=2A$.
   5. Substitute: $5=A+4A=5A\Rightarrow A=1,\,B=2$.
   6. $I=\int\dfrac{D+2D'}{D}dx=\int1\,dx+2\int\dfrac{D'}{D}dx=x+2\log|D|+c=x+2\log|\sin x-2\cos x|+c$.
   7. Comparing to the given form, $a=2$.

   Common Mistakes

   • Sign errors when matching coefficients of $\sin x$ and $\cos x$ (especially the sign of the $-2\cos x$ term inside $D$).
   • Forgetting to convert $\tan x$ to $\sin x/\cos x$ first, which is what makes the denominator match the target logarithm form $\sin x-2\cos x$.

   ✓Final answer

   The correct option is (D) — $2$.

   ANSWER: D

6. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\int e^{\sin x}\dfrac{(x\cos^3x-\sin x)}{\cos^2x}dx=$ (A) $e^{\sin x}(x-\sec x)+C$ (B) $e^{\sin x}(x-\text{cosec}\,x)+C$ (C) $e^{\sin x}(x+\sec x)+C$ (D) $e^{\sin x}(x+\text{cosec}\,x)+C$

   ›Reveal solutionSolution

   Simplifying the messy fraction to $x\cos x-\sec x\tan x$ reveals the integrand as the derivative of the product $e^{\sin x}(x-\sec x)$, making the antiderivative immediate.

   Concept and Intuition

   Integrals of the form $\int e^{g(x)}\big[h'(x)+h(x)g'(x)\big]dx=e^{g(x)}h(x)+C$ (a generalisation of $\int e^x[f(x)+f'(x)]dx=e^xf(x)+C$). Here $g(x)=\sin x$ so $g'(x)=\cos x$, and we should look for $h(x)$ such that $h'(x)+h(x)\cos x$ matches the simplified integrand.

   Step-by-Step Solution

   1. Simplify: $\dfrac{x\cos^3x-\sin x}{\cos^2x}=\dfrac{x\cos^3x}{\cos^2x}-\dfrac{\sin x}{\cos^2x}=x\cos x-\sin x\sec^2x=x\cos x-\sec x\tan x$ (using $\sin x\sec^2x=\dfrac{\sin x}{\cos^2x}=\sec x\tan x$).
   2. Guess $h(x)=x-\sec x$; check $h'(x)+h(x)\cos x = (1-\sec x\tan x)+(x-\sec x)\cos x = 1-\sec x\tan x+x\cos x-1 = x\cos x-\sec x\tan x$ — exactly matches the simplified integrand!
   3. So $\int e^{\sin x}(x\cos x-\sec x\tan x)\,dx = e^{\sin x}(x-\sec x)+C$ directly, by the identity $\int e^{g}(h'+hg')dx=e^gh$.

   Common Mistakes

   • Trying to integrate by parts term-by-term instead of recognising the "$e^{g}(h'+hg')$" pattern, leading to a much longer (and error-prone) computation.
   • Sign slip converting $\sin x/\cos^2x$ to $\sec x\tan x$ (both are positive multiples of $\sin x/\cos^2 x$, so this is actually an identity, not an approximation — worth double-checking with the definitions directly).

   ✓Final answer

   The correct option is (A) — $e^{\sin x}(x-\sec x)+C$.

   ANSWER: A

7. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\int e^x(f(x) - f'(x)) = g(x) + C$, then $\int e^x f'(x)\,dx =$ (A) $\dfrac{1}{2}\left[e^x f(x) - g(x)\right] + C$ (B) $\dfrac{1}{2}\left[e^x f(x) + g(x)\right] + C$ (C) $\dfrac{e^x f'(x) + g(x)}{2} + C$ (D) $\dfrac{1}{2}\left[e^x f(x) + e^x g(x)\right] + C$

   ›Reveal solutionSolution

   Combining the given identity with the product-rule identity $\int e^x f dx + \int e^x f' dx = e^x f(x)+C$ isolates $\int e^x f'(x)dx = \tfrac12[e^xf(x)-g(x)]+C$.

   Concept and Intuition

   Whenever you see $\int e^x(f\pm f')dx$, remember that $\dfrac{d}{dx}[e^xf(x)] = e^xf(x)+e^xf'(x)$, so $\int e^xf(x)dx + \int e^xf'(x)dx = e^xf(x)+C$ always. This gives a second equation to pair with the one given, letting us solve for either integral separately (like solving simultaneous equations).

   Step-by-Step Solution

   1. Let $I_1=\int e^xf(x)dx$ and $I_2=\int e^xf'(x)dx$.
   2. Given: $I_1 - I_2 = g(x)+C$ ... (i)
   3. Product rule identity: $I_1+I_2 = e^xf(x)+C'$ ... (ii)
   4. Subtract (i) from (ii): $(I_1+I_2)-(I_1-I_2) = 2I_2 = e^xf(x)-g(x) + (C'-C)$.
   5. So $I_2 = \dfrac{e^xf(x)-g(x)}{2} + C''$, i.e. $\int e^xf'(x)dx = \tfrac12[e^xf(x)-g(x)]+C$.

   Common Mistakes

   • Sign error when subtracting the two integral identities — carefully track which is $I_1-I_2$ and which is $I_1+I_2$.
   • Forgetting the product-rule identity entirely and trying to integrate $f'(x)$ directly without more information about $f$.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{2}\left[e^x f(x) - g(x)\right] + C$.

   ANSWER: A

8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \left(\dfrac{\log x - 1}{1+(\log x)^2}\right)^2 dx=$ (A) $\dfrac{x e^x}{1+x^2}+c$ (B) $\dfrac{x}{1+(\log x)^2}+c$ (C) $\dfrac{\log x}{(\log x)^2+1}+c$ (D) $\dfrac{x}{x^2+1}+c$

   ›Reveal solutionSolution

   The integrand is recognized as the derivative of $\dfrac{x}{1+(\log x)^2}$ via the quotient rule, so the antiderivative is that expression directly.

   Concept and Intuition

   Many integrals involving $\log x$ in a rational combination are disguised derivatives of a simple quotient $\dfrac{x}{P(\log x)}$. The strategy is to guess a plausible antiderivative form (informed by the structure of the options) and verify by differentiating.

   Step-by-Step Solution

   1. Guess $g(x) = \dfrac{x}{1+(\log x)^2}$ (matching option B's form).
   2. Differentiate using the quotient rule: $g'(x) = \dfrac{1\cdot(1+(\log x)^2) - x\cdot 2\log x \cdot \frac{1}{x}}{(1+(\log x)^2)^2}$.
   3. Simplify numerator: $(1+(\log x)^2) - 2\log x = 1 - 2\log x + (\log x)^2 = (\log x - 1)^2$.
   4. So $g'(x) = \dfrac{(\log x-1)^2}{(1+(\log x)^2)^2} = \left(\dfrac{\log x - 1}{1+(\log x)^2}\right)^2$, exactly the integrand.
   5. Therefore $\displaystyle\int \left(\dfrac{\log x - 1}{1+(\log x)^2}\right)^2 dx = \dfrac{x}{1+(\log x)^2}+c$.

   Common Mistakes

   • Attempting substitution $u=\log x$ directly without accounting for the extra factor of $x$ that appears when converting $dx$ to $du$ — the quotient-rule verification approach avoids this trap entirely.

   ✓Final answer

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

   ANSWER: B

9. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If $f(x)$ is anti derivative of $g(x)$ and $\int f(x)g(x)\left(1 + f^2(x)\right)dx = F(x)$, then $F(x) =$ (A) $\dfrac{(1 + f^2(x))^2}{4} + C$ (B) $\dfrac{(1 + f^2(x))^2}{2} + C$ (C) $\dfrac{f^2(x)g(x)}{4} + C$ (D) $\dfrac{g^2(x)f(x)}{4} + C$

   ›Reveal solutionSolution

   Substituting $u=f(x)$ turns the integral into $\int u(1+u^2)\,du$, a simple polynomial integral in $u$ that equals $\dfrac{(1+u^2)^2}{4}+C$ after absorbing a constant — giving $F(x)=\dfrac{(1+f^2(x))^2}{4}+C$.

   Concept and Intuition

   Since $f$ is an antiderivative of $g$, we have $f'(x)=g(x)$, so $g(x)\,dx=df$. This means the integral, which looks complicated in $x$, is actually a clean substitution integral in the single variable $u=f(x)$.

   Step-by-Step Solution

   1. Let $u=f(x)$, so $du=f'(x)\,dx=g(x)\,dx$.
   2. The integral becomes $\int f(x)g(x)(1+f^2(x))\,dx=\int u(1+u^2)\,du$.
   3. Expand and integrate: $\int(u+u^3)\,du=\dfrac{u^2}{2}+\dfrac{u^4}{4}+C_1$.
   4. Verify this matches $\dfrac{(1+u^2)^2}{4}$ up to a constant: $\dfrac{(1+u^2)^2}{4}=\dfrac{1+2u^2+u^4}{4}=\dfrac14+\dfrac{u^2}{2}+\dfrac{u^4}{4}$ — differs from step 3's result only by the constant $\tfrac14$, which is absorbed into $C$.
   5. So $F(x)=\dfrac{(1+f^2(x))^2}{4}+C$.
   6. Quick check by differentiating option A: $\dfrac{d}{dx}\left[\dfrac{(1+f^2)^2}{4}\right]=\dfrac{1}{4}\cdot2(1+f^2)\cdot2f\cdot f'=f\,f'(1+f^2)=f(x)g(x)(1+f^2(x))$ — exactly the original integrand, confirming the antiderivative.

   Common Mistakes

   • Forgetting that $g(x)\,dx=df$ and instead trying to integrate directly in $x$ without substitution, which obscures the simple polynomial structure.
   • Missing the constant-of-integration equivalence between $\dfrac{u^2}{2}+\dfrac{u^4}{4}$ and $\dfrac{(1+u^2)^2}{4}$ and thinking they must exactly match term-by-term (they only need to match up to an additive constant).

   ✓Final answer

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

   ANSWER: A

10. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{e^{\tan^{-1}x}}{1+x^2}\left[\left(\sec^{-1}\sqrt{1+x^2}\right)^2 + \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]dx =$ (A) $e^{\tan^{-1}x}(\tan^{-1}x)^2 + C$ (B) $e^{\tan^{-1}x}(\sec^{-1}x)^2 + C$ (C) $e^{\tan^{-1}x}\left(\sec^{-1}\left(\sqrt{1+x^2}\right)\right) + C$ (D) $e^{\tan^{-1}x}\left(\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right) + C$

    ›Reveal solutionSolution

    Simplifying the inverse-trig bracket to a function of $\tan^{-1}x$ turns the integral into the recognizable pattern $\int e^u(u^2+2u)\,du = e^u u^2+C$.

    Concept and Intuition

    Composite inverse-trig expressions like $\sec^{-1}\sqrt{1+x^2}$ and $\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)$ are classic disguises for $\tan^{-1}x$ and $2\tan^{-1}x$ (substitute $x=\tan\theta$ to see it). Once revealed, the whole integrand collapses to a substitution $u=\tan^{-1}x$, and the integral becomes a standard "$e^u$ times a function whose derivative pattern matches" form.

    Step-by-Step Solution

    1. Let $x=\tan\theta$. Then $\sqrt{1+x^2}=\sec\theta$, so $\sec^{-1}\sqrt{1+x^2}=\theta=\tan^{-1}x$.
    2. Also $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta$, so $\cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)=2\theta = 2\tan^{-1}x$.
    3. The bracket becomes $(\tan^{-1}x)^2+2\tan^{-1}x$.
    4. Let $u=\tan^{-1}x$, so $du=\dfrac{dx}{1+x^2}$. The integral becomes $\displaystyle\int e^u(u^2+2u)\,du$.
    5. Note $\dfrac{d}{du}(e^uu^2)=e^uu^2+e^u\cdot2u = e^u(u^2+2u)$ — exactly the integrand.
    6. So $\displaystyle\int e^u(u^2+2u)\,du = e^uu^2+C = e^{\tan^{-1}x}(\tan^{-1}x)^2+C$.

    Common Mistakes

    • Missing the identity $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2\tan^{-1}x$ and trying to integrate the original mess directly.
    • Forgetting that $du=dx/(1+x^2)$ already accounts for the $\frac{1}{1+x^2}$ factor out front — don't integrate it again.

    ✓Final answer

    The correct option is (A) — $e^{\tan^{-1}x}(\tan^{-1}x)^2 + C$.

    ANSWER: A

11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{1 + x\cos x}{x\left[1 - x^2(e^{\sin x})^2\right]}\, dx =$ (A) $\dfrac{1}{2}\log\left|\dfrac{(xe^{\sin x})^2}{(xe^{\sin x})^2 + 1}\right| + c$ (B) $-\dfrac{1}{2}\log\left|\dfrac{(xe^{\sin x})^2}{(xe^{\sin x})^2 + 1}\right| + c$ (C) $\dfrac{1}{2}\log\left|\dfrac{(xe^{\sin x})^2}{(xe^{\sin x})^2 - 1}\right| + c$ (D) $-\dfrac{1}{2}\log\left|\dfrac{(xe^{\sin x})^2}{(xe^{\sin x})^2 - 1}\right| + c$

    ›Reveal solutionSolution

    Substituting $u=xe^{\sin x}$ (whose derivative matches the numerator) reduces the integral to a standard partial-fractions form.

    Concept and Intuition

    The numerator $1+x\cos x$ is recognizable as (up to a factor $e^{\sin x}$) the derivative of $xe^{\sin x}$; spotting this collapses the whole integrand into a rational function of $u=xe^{\sin x}$.

    Step-by-Step Solution

    1. Let $u=xe^{\sin x}$. Then $\dfrac{du}{dx}=e^{\sin x}+xe^{\sin x}\cos x=e^{\sin x}(1+x\cos x)$.
    2. So $(1+x\cos x)\,dx=\dfrac{du}{e^{\sin x}}=\dfrac{x}{u}\,du$ (using $e^{\sin x}=u/x$).
    3. Original integral $=\displaystyle\int\dfrac{(1+x\cos x)\,dx}{x(1-x^2e^{2\sin x})}=\int\dfrac{(x/u)\,du}{x(1-u^2)}=\int\dfrac{du}{u(1-u^2)}$.
    4. Partial fractions: $\dfrac{1}{u(1-u)(1+u)}=\dfrac1u+\dfrac{1/2}{1-u}-\dfrac{1/2}{1+u}$.
    5. Integrate: $\log|u|-\tfrac12\log|1-u|-\tfrac12\log|1+u|+C=\log|u|-\tfrac12\log|1-u^2|+C$.
    6. Combine logs: $=\tfrac12\log\left|\dfrac{u^2}{1-u^2}\right|+C$, and since $|1-u^2|=|u^2-1|$, this equals $\tfrac12\log\left|\dfrac{u^2}{u^2-1}\right|+C$.
    7. Substitute back $u=xe^{\sin x}$.

    Common Mistakes

    • Missing the substitution and attempting to integrate directly.
    • Sign confusion between $1-u^2$ and $u^2-1$ (resolved by the absolute value).

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{2}\log\left|\dfrac{(xe^{\sin x})^2}{(xe^{\sin x})^2 - 1}\right| + c$.

    ANSWER: C

12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{(x-1)^2}{(x^2+1)^2}\,dx = \tan^{-1}(x) + g(x) + k$, then $g(x)$ is equal to ____ (A) $\tan^{-1}\left(\dfrac{x}{2}\right)$ (B) $\dfrac{1}{x^2+1}$ (C) $\dfrac{1}{2(x^2+1)}$ (D) $\dfrac{2}{x^2+1}$

    ›Reveal solutionSolution

    Splitting $(x-1)^2=(x^2+1)-2x$ turns the integral into a standard $\tan^{-1}$ piece plus an easy substitution piece, giving $g(x)=\dfrac{1}{x^2+1}$.

    Concept and Intuition

    Whenever a numerator can be rewritten using the denominator's structure (here $x^2+1$), splitting the fraction usually reduces the problem to two standard integrals.

    Step-by-Step Solution

    1. Expand the numerator: $(x-1)^2=x^2-2x+1=(x^2+1)-2x$.
    2. Split the integral:

    $\int\frac{(x-1)^2}{(x^2+1)^2}dx=\int\frac{(x^2+1)-2x}{(x^2+1)^2}dx=\int\frac{1}{x^2+1}dx-\int\frac{2x}{(x^2+1)^2}dx.$

    3. First piece: $\displaystyle\int\frac{dx}{x^2+1}=\tan^{-1}x$.
    4. Second piece: let $u=x^2+1$, $du=2x\,dx$: $\displaystyle\int\frac{2x\,dx}{(x^2+1)^2}=\int\frac{du}{u^2}=-\frac1u=-\frac{1}{x^2+1}$.
    5. Combine: $\tan^{-1}x-\left(-\dfrac{1}{x^2+1}\right)+c=\tan^{-1}x+\dfrac{1}{x^2+1}+c$.
    6. Matching against $\tan^{-1}x+g(x)+k$: $g(x)=\dfrac{1}{x^2+1}$.

    Common Mistakes

    • Sign error on the second integral (the minus signs can cancel incorrectly if not careful).
    • Forgetting to subtract the derivative-of-denominator piece separately.

    ✓Final answer

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

    ANSWER: B