Q.Integrate the following function: $\int (4 e^{3x} + 1) dx$

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{e^{2x}}{\sqrt[4]{e^x+1}} \, dx =$ (A) $\dfrac{4}{7}(e^x+1)^{4/3}(3e^x-1)+c$ (B) $\dfrac{2}{21}(e^x+1)^{3/4}(3e^x-7)+c$ (C) $\dfrac{4}{21}(e^x+1)^{3/4}(3e^x-4)+c$ (D) $\dfrac{8}{21}(e^x+1)^{3/4}(3e^x-1)+c$

   ›Reveal solutionSolution

   Substitute $u=e^x+1$ to turn this into a simple power-rule integral; simplifies to $\dfrac4{21}(e^x+1)^{3/4}(3e^x-4)+c$ — (C).

   Concept and Intuition

   Writing $e^{2x}=e^x\cdot e^x$ and using $u=e^x+1$ (so $e^x=u-1$, $du=e^xdx$) converts the whole integral into a polynomial-in-$u$ times a fractional power of $u$ — directly integrable term by term.

   Step-by-Step Solution

   1. Let $u=e^x+1\Rightarrow du=e^x\,dx,\ e^x=u-1$.
   2. $e^{2x}dx = e^x\cdot(e^x dx) = (u-1)\,du$.
   3. Integral: $\displaystyle\int\dfrac{(u-1)\,du}{u^{1/4}}=\int\left(u^{3/4}-u^{-1/4}\right)du = \dfrac{u^{7/4}}{7/4}-\dfrac{u^{3/4}}{3/4}+c=\dfrac47u^{7/4}-\dfrac43u^{3/4}+c$.
   4. Factor out $u^{3/4}$: $u^{3/4}\left(\dfrac47u-\dfrac43\right)+c$. Common denominator $21$: $\dfrac47u=\dfrac{12u}{21}$, $\dfrac43=\dfrac{28}{21}$, so this is $u^{3/4}\cdot\dfrac{12u-28}{21}=\dfrac{4}{21}u^{3/4}(3u-7)+c$. …

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2. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int \left(\sum_{r=0}^{\infty} \dfrac{x^r 3^r}{r!}\right) dx =$ (A) $e^x + c$ (B) $\dfrac{e^{3x}}{3} + c$ (C) $3e^{3x} + c$ (D) $3e^x + c$

   ›Reveal solutionSolution

   $\sum_{r=0}^\infty \dfrac{(3x)^r}{r!}=e^{3x}$, so the integral is $\dfrac{e^{3x}}{3}+c$.

   Concept and Intuition

   Recognising the exponential series $\sum x^r/r! = e^x$ turns an infinite-series integral into a one-line exponential integration.

   Step-by-Step Solution

   1. $\sum_{r=0}^\infty \dfrac{x^r3^r}{r!} = \sum_{r=0}^\infty \dfrac{(3x)^r}{r!} = e^{3x}$.
   2. $\int e^{3x}\,dx = \dfrac{e^{3x}}{3}+c$.

   Common Mistakes …

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3. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.$\int\left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right)dx=$ (A) $\frac{2^x}{\log 2}+4x+C$ (B) $8\cdot\frac{2^x}{\log 2}-4x+C$ (C) $8\cdot\frac{2^x}{\log 2}+4x+C$ (D) $\frac{2^x}{\log 2}-4x+C$

   ›Reveal solutionSolution

   Rewrite everything as powers of 2 and simplify the fraction before integrating; the answer is $8\cdot\frac{2^x}{\log 2}+4x+C$.

   Concept and Intuition

   Exponential integrands often simplify dramatically once every term is expressed with the same base. Here base 2 works for 8, 4 and $2^{2x}$ alike, turning a scary-looking fraction into a sum of a simple exponential and a constant.

   Step-by-Step Solution

   1. $8^{1+x}=8\cdot 8^x=8\cdot(2^3)^x=8\cdot 2^{3x}$.
   2. $4^{1+x}=4\cdot 4^x=4\cdot(2^2)^x=4\cdot2^{2x}$.
   3. Divide both by $2^{2x}$: $\dfrac{8\cdot2^{3x}}{2^{2x}}+\dfrac{4\cdot2^{2x}}{2^{2x}}=8\cdot2^{x}+4$.
   4. Integrate term by term: $\int 8\cdot2^x\,dx=8\cdot\dfrac{2^x}{\log2}$, and $\int 4\,dx=4x$. …

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4. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\left(x + \sqrt{x}\right)dx = e^{\sqrt{x}}\left[Ax + B\sqrt{x} + C\right] + K$, then $A + B + C =$ (A) -2 (B) 2 (C) 4 (D) -4

   ›Reveal solutionSolution

   Substituting $t=\sqrt x$ converts the integral into a standard $\int e^t(t^2+t)\,dt$, which evaluates to $e^{\sqrt x}(2x-2\sqrt x+2)+K$, so $A+B+C=2-2+2=2$.

   Concept and Intuition

   Integrals with $e^{\sqrt x}$ and $\sqrt x$ everywhere are cleaned up by the substitution $t=\sqrt x$, turning them into polynomial-times-exponential integrals that use the standard reduction formula $\int t^ne^t\,dt$.

   Step-by-Step Solution

   1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
   2. Integral: $\displaystyle\int\dfrac{e^t}{t}(t^2+t)\cdot2t\,dt = \int e^t\cdot t(t+1)\cdot2\,dt = 2\int e^t(t^2+t)\,dt$.
   3. Use $\int t^2e^t\,dt = e^t(t^2-2t+2)$ and $\int te^t\,dt=e^t(t-1)$ (both by repeated integration by parts).
   4. Sum: $e^t(t^2-2t+2)+e^t(t-1) = e^t(t^2-t+1)$.
   5. Multiply by $2$: $2e^t(t^2-t+1) = e^t(2t^2-2t+2)$. …

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5. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{3e^{x} - 7e^{-x}}{7e^{x} + 3e^{-x}}\,dx = Kx + L\log\left(e^{-2x} + \dfrac{7}{3}\right) + C$, then $K + L =$ (A) $\dfrac{-3}{38}$ (B) $\dfrac{21}{38}$ (C) $\dfrac{38}{21}$ (D) $\dfrac{-38}{3}$

   ›Reveal solutionSolution

   Splitting the numerator as a combination of the denominator and its derivative gives $Kx+L\log D$; converting $D$ to the $e^{-2x}+7/3$ form shows $K+L=\dfrac{38}{21}$.

   Concept and Intuition

   For an integral of the form $\int\dfrac{N(x)}{D(x)}dx$ where $N,D$ are combinations of $e^x,e^{-x}$, the standard trick is to write $N = A\,D + B\,D'$. Then $\int\dfrac{N}{D}dx = A\int dx + B\int\dfrac{D'}{D}dx = Ax+B\log|D|+C$ — turning a messy rational-exponential integral into an elementary one.

   Step-by-Step Solution

   1. Let $D = 7e^x+3e^{-x}$, so $D' = 7e^x-3e^{-x}$.
   2. Seek $A,B$ with $3e^x-7e^{-x} = A(7e^x+3e^{-x})+B(7e^x-3e^{-x})$.
   3. Coefficient of $e^x$: $7A+7B=3$. Coefficient of $e^{-x}$: $3A-3B=-7$.
   4. Solve: $A+B=\dfrac37$, $A-B=-\dfrac73$. Adding: $2A = \dfrac37-\dfrac73=\dfrac{9-49}{21}=-\dfrac{40}{21}\Rightarrow A=-\dfrac{20}{21}$. Subtracting: $2B=\dfrac37+\dfrac73=\dfrac{9+49}{21}=\dfrac{58}{21}\Rightarrow B=\dfrac{29}{21}$.
   5. So $\int\frac{N}{D}dx = Ax + B\log D + C$, i.e. using $D=7e^x+3e^{-x}=e^x\left(7+3e^{-2x}\right)=3e^x\left(e^{-2x}+\tfrac73\right)$: $\log D = x+\log 3+\log\left(e^{-2x}+\tfrac73\right)$. …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int 4^{x}\,4^{4^{x}}\,4^{4^{4^{x}}}\,dx = A\,4^{4^{4^{x}}}+c$, then $A=$ (A) $\dfrac{1}{\ln 4}$ (B) $\dfrac{1}{(\ln 4)^2}$ (C) $\dfrac{1}{(\ln 4)^3}$ (D) $\dfrac{1}{(\ln 4)^4}$

   ›Reveal solutionSolution

   Repeated application of $\frac{d}{dx}a^{g(x)}=a^{g(x)}\log a\cdot g'(x)$ through a triple exponential tower produces a factor of $(\ln4)^3$, so integrating backwards needs $A=1/(\ln4)^3$.

   Concept and Intuition

   For a tower of exponentials $4^{4^{4^x}}$, differentiating peels one exponential at a time, each time multiplying by $\ln4$ and by the derivative of the next inner exponential. Because the given integrand is exactly the product $4^x\cdot4^{4^x}\cdot4^{4^{4^x}}$ — the "chain" of factors produced by differentiating the tower — this is a reverse-chain-rule integral: recognizing the derivative of the tower directly gives the antiderivative.

   Step-by-Step Solution

   1. Let $F(x)=4^{4^{4^x}}$. Differentiate using the chain rule from the outside in:

   $F'(x)=4^{4^{4^x}}\ln4\cdot\frac{d}{dx}\left(4^{4^x}\right).$

   2. Now $\dfrac{d}{dx}\left(4^{4^x}\right)=4^{4^x}\ln4\cdot\dfrac{d}{dx}(4^x)=4^{4^x}\ln4\cdot 4^x\ln4 = 4^x\cdot4^{4^x}\cdot(\ln4)^2$.
   3. Substituting back: $F'(x)= (\ln4)^3\cdot 4^x\cdot4^{4^x}\cdot4^{4^{4^x}}$. …

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7. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{a^{x+1}}{a^x + a^{-x}}\, dx = A\log(a^x + a^{-x}) + Bx + C$ then $\dfrac{A}{B} =$ (A) $\log_a e$ (B) $\log_e a$ (C) $\dfrac{a}{\log_e a}$ (D) $\dfrac{a}{\log_a e}$

   ›Reveal solutionSolution

   Simplifying via $u=a^{2x}+1$ and re-expressing in terms of $a^x+a^{-x}$ gives $A/B = \log_a e$.

   Concept and Intuition

   The trick with $\int \dfrac{a^{x+1}}{a^x+a^{-x}}dx$ is to multiply through by $a^x/a^x$ so the denominator becomes a polynomial in $a^{2x}$, turning it into a standard $\int \dfrac{du}{u}$ form after substitution. The subtlety is that this substitution naturally produces $\ln(a^{2x}+1)$, which must be related back to the given target form $\ln(a^x+a^{-x})$ using the identity $a^{2x}+1 = a^x(a^x+a^{-x})$.

   Step-by-Step Solution

   1. Multiply numerator and denominator by $a^x$: $\dfrac{a\cdot a^x}{a^x+a^{-x}} = \dfrac{a\cdot a^{2x}}{a^{2x}+1}$.
   2. Let $u = a^{2x}+1$, so $du = 2a^{2x}\ln a\,dx$, i.e. $a^{2x}dx = \dfrac{du}{2\ln a}$.
   3. Integral becomes $a\int \dfrac{1}{u}\cdot\dfrac{du}{2\ln a} = \dfrac{a}{2\ln a}\ln|u| + C = \dfrac{a}{2\ln a}\ln(a^{2x}+1)+C$.
   4. Use $a^{2x}+1 = a^x(a^x+a^{-x})$, so $\ln(a^{2x}+1) = \ln(a^x+a^{-x}) + x\ln a$. …

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8. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} e^{\tan^2\theta} \sin^2\theta \tan\theta\, d\theta =$ (A) $\dfrac{1}{2}\left(\dfrac{e}{2} - 1\right)$ (B) $\dfrac{e}{2} - 1$ (C) $\dfrac{\pi}{2}$ (D) $2\left(\dfrac{\pi}{2} - e\right)$

   ›Reveal solutionSolution

   Substituting $v=\tan^2\theta$ converts the integral into $\frac12\int_0^1 \frac{ve^v}{(1+v)^2}\,dv$, which is exactly a product-rule derivative of $\frac{e^v}{1+v}$, giving the clean closed form $\frac12\left(\frac{e}{2}-1\right)$.

   Concept and Intuition

   When an integral mixes $\tan\theta$, $\sec^2\theta$-type factors, and an exponential of $\tan^2\theta$, substituting $v=\tan^2\theta$ (rather than $u=\tan\theta$) often collapses the trigonometric parts neatly, because $dv$ naturally pulls out a $\tan\theta\sec^2\theta$ factor that matches what's needed.

   Step-by-Step Solution

   1. Let $v=\tan^2\theta$. Then $dv = 2\tan\theta\sec^2\theta\,d\theta = 2\tan\theta(1+\tan^2\theta)\,d\theta = 2\tan\theta(1+v)\,d\theta$.
   2. So $\tan\theta\,d\theta = \dfrac{dv}{2(1+v)}$.
   3. Also $\sin^2\theta = \tan^2\theta\cos^2\theta = \dfrac{\tan^2\theta}{1+\tan^2\theta} = \dfrac{v}{1+v}$.
   4. The integrand $e^{\tan^2\theta}\sin^2\theta\tan\theta\,d\theta$ becomes $e^v\cdot\dfrac{v}{1+v}\cdot\dfrac{dv}{2(1+v)} = \dfrac{v\,e^v}{2(1+v)^2}\,dv$.
   5. Limits: as $\theta:0\to\pi/4$, $v=\tan^2\theta: 0\to1$.
   6. So the integral is $\dfrac12\displaystyle\int_0^1 \dfrac{v e^v}{(1+v)^2}\,dv$.
   7. Write $v = (1+v)-1$: $\dfrac{v}{(1+v)^2} = \dfrac1{1+v}-\dfrac1{(1+v)^2}$. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int\left(\sum_{r=0}^{\infty} \dfrac{x^r 2^r}{r!}\right)dx =$ (A) $e^x + c$ (B) $\dfrac{-2}{1-2x}+c$ (C) $2e^{2x}+c$ (D) $\dfrac{e^{2x}}{2}+c$

   ›Reveal solutionSolution

   Recognising the Maclaurin series of $e^{2x}$ inside the sum turns this into a one-line integral, giving $\dfrac{e^{2x}}{2}+c$.

   Concept and Intuition

   The exponential series $e^{y}=\sum_{r=0}^\infty \dfrac{y^r}{r!}$ is one of the most important series to recognise instantly. Here $y=2x$, so the given sum is exactly $e^{2x}$ in disguise — the whole problem reduces to a basic integral once you see this.

   Step-by-Step Solution

   1. $\displaystyle\sum_{r=0}^\infty \dfrac{x^r 2^r}{r!} = \sum_{r=0}^\infty \dfrac{(2x)^r}{r!} = e^{2x}$.
   2. So the integral becomes $\displaystyle\int e^{2x}\,dx$.
   3. $\displaystyle\int e^{2x}\,dx = \dfrac{e^{2x}}{2}+c$. …

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10. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.$\displaystyle\int e^{2x+3}\sin 6x\,dx=$ (A) $\dfrac{e^{2x+3}}{40}(2\sin 6x+6\cos 6x)+c$ (B) $\dfrac{e^{2x+3}}{40}(2\cos 6x+6\sin 6x)+c$ (C) $\dfrac{e^{2x+3}}{20}(\sin 6x-3\cos 6x)+c$ (D) $\dfrac{e^{2x+3}}{20}(\cos 6x-3\sin 6x)+c$

    ›Reveal solutionSolution

    A direct application of the standard $\int e^{ax}\sin bx\,dx$ reduction formula, treating the constant $e^3$ as absorbed back into $e^{2x+3}$. The answer is $\dfrac{e^{2x+3}}{20}(\sin6x-3\cos6x)+c$.

    Concept and Intuition

    The classic formula $\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+c$ (derived via two rounds of integration by parts) applies here with $a=2$, $b=6$; the extra constant factor $e^3$ in $e^{2x+3}=e^3e^{2x}$ just rides along and recombines at the end.

    Step-by-Step Solution

    1. Write $e^{2x+3}=e^3\cdot e^{2x}$, so $\displaystyle\int e^{2x+3}\sin6x\,dx=e^3\int e^{2x}\sin6x\,dx$.
    2. Apply the formula with $a=2,b=6$: $a^2+b^2=4+36=40$.
    3. $\displaystyle\int e^{2x}\sin6x\,dx=\dfrac{e^{2x}}{40}(2\sin6x-6\cos6x)+c$.
    4. Multiply back by $e^3$: $e^3\cdot\dfrac{e^{2x}}{40}(2\sin6x-6\cos6x)=\dfrac{e^{2x+3}}{40}(2\sin6x-6\cos6x)$. …

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11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.$\displaystyle\int_0^1 (\sqrt{10})^{2x}dx =$ (A) $\dfrac{10}{\log 10}$ (B) $\dfrac{9}{\log 10}$ (C) $\dfrac{1}{\log 10}$ (D) $\dfrac{4}{\log 5}$

    ›Reveal solutionSolution

    Simplify the exponential base first ($(\sqrt{10})^{2x}=10^x$), then integrate directly using the standard exponential-integral formula.

    Concept and Intuition

    $\displaystyle\int a^x\,dx = \dfrac{a^x}{\ln a}+C$ for $a>0,a\neq1$. Recognizing $(\sqrt{10})^{2x}=10^x$ first avoids working with an awkward fractional-power base.

    Step-by-Step Solution

    1. $(\sqrt{10})^{2x} = \left(10^{1/2}\right)^{2x} = 10^x$.
    2. $\displaystyle\int_0^1 10^x\,dx = \left[\dfrac{10^x}{\ln10}\right]_0^1 = \dfrac{10^1-10^0}{\ln10} = \dfrac{10-1}{\ln10} = \dfrac{9}{\ln10}$.

    Common Mistakes …

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12. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.$\displaystyle\int_{\alpha-1}^{\alpha} \dfrac{e^x(\alpha - x)}{(x-\alpha+1)^2}\,dx =$ (A) $2e^{\alpha} + e$ (B) $\dfrac{2e^{\alpha+2}}{e-2}$ (C) $e^{\alpha}\dfrac{(e+2)}{2}$ (D) $e^{\alpha}\left(\dfrac{e-2}{2}\right)$

    ›Reveal solutionSolution

    This is a shifted $e^x[f(x)+f'(x)]$ integral; substituting $t=x-\alpha$ turns it into the standard $\int \dfrac{t\,e^t}{(t+1)^2}dt = \dfrac{e^t}{t+1}+C$ pattern, evaluated over $t\in[0,1]$.

    Concept and Intuition

    Many definite integrals involving $e^x$ divided by a squared linear factor are disguised applications of the identity $\int e^x\big[g(x)+g'(x)\big]dx = e^x g(x)+C$, which follows straight from the product rule. Recognizing the pattern (a term like $\dfrac{1}{t+1}$ together with its derivative $-\dfrac{1}{(t+1)^2}$, glued together by $e^t$) avoids messy direct integration.

    Step-by-Step Solution

    1. Shift variable: let $t=x-\alpha$, so $x=\alpha+t$, and the range $x:\alpha\to\alpha+1$ becomes $t:0\to1$; the factor $x-\alpha+1$ becomes $t+1$.
    2. The integral becomes $e^\alpha\displaystyle\int_0^1 \dfrac{t\,e^t}{(t+1)^2}\,dt$.
    3. Write $t=(t+1)-1$, so $\dfrac{t}{(t+1)^2}=\dfrac{1}{t+1}-\dfrac{1}{(t+1)^2}$.
    4. Note $\dfrac{d}{dt}\!\left[\dfrac{e^t}{t+1}\right]=\dfrac{e^t(t+1)-e^t}{(t+1)^2}=\dfrac{e^t\,t}{(t+1)^2}$ — exactly the integrand. …

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