Q.Integrate the following function: $\sin^3 (2x + 1)$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{x^4 - 1}{x^2\sqrt{x^4 + x^2 + 1}}dx =$ (A) $\frac{2\sqrt{x^4+x^2+1}}{x} + c$ (B) $\frac{\sqrt{x^4+x^2+1}}{x} + c$ (C) $\frac{\sqrt{x^4+x^2+1}}{2x} + c$ (D) $\frac{4\sqrt{x^4+x^2+1}}{x} + c$

   ›Reveal solutionSolution

   Differentiating the candidate $\frac{\sqrt{x^4+x^2+1}}{x}$ reproduces the given integrand exactly, confirming it as the antiderivative.

   Concept and Intuition

   When an integrand looks like it could come from a quotient rule (a square root over a power of $x$), it is often faster to differentiate a plausible candidate of that shape and check, rather than search for a substitution from scratch.

   Step-by-Step Solution

   1. Try $g(x)=\dfrac{\sqrt{x^4+x^2+1}}{x}=\dfrac{N}{x}$ where $N=\sqrt{x^4+x^2+1}$.
   2. $N'=\dfrac{4x^3+2x}{2\sqrt{x^4+x^2+1}}=\dfrac{x(2x^2+1)}{N}$.
   3. Quotient rule: $g'(x)=\dfrac{N'x-N}{x^2}=\dfrac{\frac{x^2(2x^2+1)}{N}-N}{x^2}=\dfrac{x^2(2x^2+1)-N^2}{Nx^2}$.
   4. $N^2=x^4+x^2+1$, so the numerator is $x^2(2x^2+1)-(x^4+x^2+1)=2x^4+x^2-x^4-x^2-1=x^4-1$.
   5. So $g'(x)=\dfrac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$ — exactly the given integrand. …

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2. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin x}{\sin 3x}\, dx =$ (A) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (B) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$ (C) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (D) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$

   ›Reveal solutionSolution

   Use the triple-angle identity to cancel $\sin x$, rewrite in terms of $\cos2x$, then apply the Weierstrass-type substitution $t=\tan x$ to reduce to a standard rational integral. The answer is (A).

   Concept and Intuition

   $\sin3x$ factors as $\sin x(3-4\sin^2x)$, so $\sin x$ cancels immediately with the numerator, turning a trigonometric-looking integral into a much simpler one in $\sin^2x$ (hence in $\cos2x$), which is a textbook target for the $t=\tan x$ substitution.

   Step-by-Step Solution

   1. $\sin3x = 3\sin x - 4\sin^3x = \sin x(3-4\sin^2x)$.
   2. $\dfrac{\sin x}{\sin3x} = \dfrac{1}{3-4\sin^2x}$.
   3. Using $\sin^2x=\dfrac{1-\cos2x}2$: $4\sin^2x = 2-2\cos2x$, so $3-4\sin^2x = 1+2\cos2x$.
   4. Integral becomes $\displaystyle\int\dfrac{dx}{1+2\cos2x}$.
   5. Substitute $t=\tan x$, $dx=\dfrac{dt}{1+t^2}$, $\cos2x=\dfrac{1-t^2}{1+t^2}$: $1+2\cos2x = \dfrac{(1+t^2)+2(1-t^2)}{1+t^2} = \dfrac{3-t^2}{1+t^2}$.
   6. Integral $= \displaystyle\int \dfrac{1+t^2}{3-t^2}\cdot\dfrac{dt}{1+t^2} = \int\dfrac{dt}{3-t^2}$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$ (A) $-\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (B) $\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (C) $\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$ (D) $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$

   ›Reveal solutionSolution

   Factor $\cos^3x$ using $\cos^2x=(1-\sin x)(1+\sin x)$ and substitute $t=\sin x$; the resulting antiderivative can equivalently be written in the $\cos^4x/(1+\sin x)^4$ form given in the options (they differ only by an added constant). Answer: $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$.

   Concept and Intuition

   Integrals of $\cos^{\text{odd}}x$ over powers of $(1+\sin x)$ are handled by peeling off one factor of $\cos x$ to pair with $dx$ (making $d(\sin x)$) and expressing the remaining even power of $\cos x$ in terms of $\sin x$. Since the MCQ options are phrased in terms of $\cos^4x$ rather than $\sin x$ directly, it is often faster (and safer against sign traps) to guess-and-check an antiderivative of that shape by differentiating a general form $A\cos^4x(1+\sin x)^{-n}$ and matching powers/coefficients — this is exactly how the printed option is confirmed.

   Step-by-Step Solution

   1. $\cos^3x=\cos x\cdot\cos^2x=\cos x(1-\sin^2x)=\cos x(1-\sin x)(1+\sin x)$.
   2. Integrand $=\dfrac{\cos x(1-\sin x)(1+\sin x)}{(1+\sin x)^4}=\dfrac{\cos x(1-\sin x)}{(1+\sin x)^3}$.
   3. Let $t=\sin x$, $dt=\cos x\,dx$: $I=\displaystyle\int\frac{1-t}{(1+t)^3}dt$. Writing $1-t=2-(1+t)$: $I=\displaystyle\int\left(\frac{2}{(1+t)^3}-\frac{1}{(1+t)^2}\right)dt=-\frac{1}{(1+t)^2}+\frac{1}{1+t}+c=\frac{t}{(1+t)^2}+c$.
   4. So $I=\dfrac{\sin x}{(1+\sin x)^2}+c$ is one valid closed form. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$ (A) $\dfrac{-(1+3\sqrt{\tan x})}{(3+\tan^2 x)^3} + C$ (B) $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$ (C) $\dfrac{-(1+\sqrt{\tan x})}{3(1+3\sqrt{\tan x})^2} + C$ (D) $\dfrac{1}{(1+3\sqrt{\tan x})^3} + C$

   ›Reveal solutionSolution

   Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Concept and Intuition

   The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

   Step-by-Step Solution

   1. Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$, matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$.
   2. So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$.
   3. Factor out $\sqrt{\cos x}$: $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$, so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$.
   4. Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$. Let $t=\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$.
   5. Let $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$.
   6. Write $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$. …

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5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3 x \cos(x-\alpha)}} =$ (A) $\dfrac{1}{\cos\alpha}\sqrt{\cot x+\tan\alpha}+c$ (B) $\dfrac{1}{\sqrt{\cos\alpha}}\sqrt{\cot x-\tan\alpha}+c$ (C) $\dfrac{-1}{\sqrt{\sin\alpha}}\sqrt{\cot x+\tan\alpha}+c$ (D) $\dfrac{-2}{\sqrt{\cos\alpha}}\sqrt{\cot x+\tan\alpha}+c$

   ›Reveal solutionSolution

   Expanding $\cos(x-\alpha)$ and substituting $u=\cot x$ reduces this to a simple square-root integral; final answer is (D).

   Concept and Intuition

   The key move is expanding $\cos(x-\alpha)$ and factoring out $\sin^4x$ from inside the square root so that a substitution $u=\cot x$ (whose differential is $-\csc^2x\,dx$, conveniently matching what's left outside) linearizes the whole thing.

   Step-by-Step Solution

   1. $\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha$.
   2. $\sin^3x\cos(x-\alpha)=\sin^3x\cos x\cos\alpha+\sin^4x\sin\alpha = \sin^4x\left(\dfrac{\cos x}{\sin x}\cos\alpha+\sin\alpha\right)=\sin^4x(\cot x\cos\alpha+\sin\alpha)$.
   3. So $\sqrt{\sin^3x\cos(x-\alpha)}=\sin^2x\sqrt{\cot x\cos\alpha+\sin\alpha}$ (taking $\sin^2x>0$ outside the root).
   4. The integral is $\displaystyle\int\dfrac{dx}{\sin^2x\sqrt{\cos\alpha\cot x+\sin\alpha}}=\int\dfrac{\csc^2x\,dx}{\sqrt{\cos\alpha\cot x+\sin\alpha}}$.
   5. Let $u=\cot x$, $du=-\csc^2x\,dx$. Integral becomes $-\displaystyle\int\dfrac{du}{\sqrt{\cos\alpha\,u+\sin\alpha}}=-\dfrac{2\sqrt{\cos\alpha\,u+\sin\alpha}}{\cos\alpha}+c$. …

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6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int e^{\sin^3 x}(\sin^8 x + 2\sin^5 x)\cos x\, dx = \dfrac{1}{3} e^{\sin^3 x} f(x) + c$, then $f(x) =$ (A) $\sin^8 x$ (B) $\sin^6 x$ (C) $\cos^8 x$ (D) $\cos^6 x$

   ›Reveal solutionSolution

   Substituting $t=\sin x$ and testing $f(t)=t^6$ against the target derivative confirms $f(x)=\sin^6x$.

   Concept and Intuition

   When an integral has the form $e^{g(x)}\cdot(\text{stuff})\cdot g'(x)$, substituting $u=g(x)$ turns it into $\int e^u\cdot h(u)\,du$ for a polynomial $h$. Matching the answer's assumed shape $\tfrac13e^{u}f$ and differentiating (product rule, since both $e^u$ and $f$ depend on $u$) lets us solve for $f$ by comparing polynomial coefficients — much safer than guessing an antiderivative by inspection alone.

   Step-by-Step Solution

   1. Let $t=\sin x$, so $dt=\cos x\,dx$. The integral becomes $\displaystyle\int e^{t^3}(t^8+2t^5)\,dt$.
   2. We're told this equals $\dfrac13 e^{t^3}f(t)+c$. Differentiate the RHS w.r.t. $t$: $\dfrac{d}{dt}\left[\dfrac13e^{t^3}f(t)\right] = \dfrac13\left[3t^2e^{t^3}f(t) + e^{t^3}f'(t)\right] = e^{t^3}\left[t^2f(t)+\dfrac13f'(t)\right]$.
   3. This must equal $e^{t^3}(t^8+2t^5)$, so $t^2f(t) + \dfrac13f'(t) = t^8+2t^5$. …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$. …

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8. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \sqrt{x}(1-x^3)^{\frac{-1}{2}}dx=\dfrac{2}{3}g(f(x))+c$, then (A) $f(x)=\sqrt{x},\ g(x)=\sin^{-1}x$ (B) $f(x)=x^{3/2},\ g(x)=\sin^{-1}x$ (C) $f(x)=x^{3/2},\ g(x)=\cos^{-1}x$ (D) $f(x)=\sqrt{x},\ g(x)=\cos^{-1}x$

   ›Reveal solutionSolution

   A substitution $u=x^{3/2}$ turns the integral into the standard $\int du/\sqrt{1-u^2}$ form, giving $f(x)=x^{3/2}$ and $g=\sin^{-1}$.

   Concept and Intuition

   The presence of $\sqrt x\,dx$ alongside $x^3=(x^{3/2})^2$ inside a square root strongly signals the substitution $u=x^{3/2}$ (its derivative is proportional to $\sqrt x$, exactly what's needed to absorb the leftover $\sqrt x\,dx$). Once substituted, the integral collapses to the standard arcsine form.

   Step-by-Step Solution

   1. Let $u=x^{3/2}$. Then $du=\dfrac32 x^{1/2}\,dx=\dfrac32\sqrt x\,dx$, so $\sqrt x\,dx=\dfrac23\,du$.
   2. Also, $u^2=x^3$, so $1-x^3=1-u^2$.
   3. Substitute into the integral: $\displaystyle\int\sqrt x\,(1-x^3)^{-1/2}\,dx=\int\dfrac23\cdot\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\int\dfrac{du}{\sqrt{1-u^2}}$.
   4. This is the standard form: $\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c$.
   5. So the integral $=\dfrac23\sin^{-1}(u)+c=\dfrac23\sin^{-1}\left(x^{3/2}\right)+c$. …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos x - \sin x}{10 + \sin 2x}\, dx =$ (A) $\dfrac{1}{2}\log(10 + \sin 2x) + c$ (B) $\dfrac{1}{3}\log(10 + \sin 2x) + c$ (C) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$ (D) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}(10 + \sin 2x) + c$

   ›Reveal solutionSolution

   The numerator $\cos x-\sin x$ is exactly $d(\sin x+\cos x)$, and the denominator rewrites in terms of $u=\sin x+\cos x$ via $\sin2x=u^2-1$. That collapses the integral to a standard $\int du/(a^2+u^2)$ arctangent form. Answer: $\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

   Concept and Intuition

   Whenever an integrand contains both $\sin x-\cos x$ (or $\cos x-\sin x$) and $\sin2x$, it is worth trying $u=\sin x+\cos x$ (or $\sin x-\cos x$) as the substitution, because $u^2=1\pm\sin2x$ links the two.

   Step-by-Step Solution

   1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$ — this is exactly the numerator times $dx$.
   2. Also $u^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin2x$, so $\sin2x=u^2-1$.
   3. Denominator: $10+\sin2x=10+u^2-1=9+u^2$.
   4. The integral becomes $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$. …

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10. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.$\int \cos\sqrt{x}\, dx =$ (A) $2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + c$ (B) $2\sqrt{x}\sin\sqrt{x} + 2\sin\sqrt{x} + c$ (C) $2\sqrt{x}\sin\sqrt{x} - 2\cos\sqrt{x} + c$ (D) $\sqrt{x}\cos\sqrt{x} - 2\sin\sqrt{x} + c$

    ›Reveal solutionSolution

    Substitute $t=\sqrt x$ to turn the integral into a standard integration-by-parts problem. Answer: $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

    Concept and Intuition

    Whenever you see $\sqrt x$ trapped inside a trig or exponential function, substituting $t=\sqrt x$ converts it into a polynomial-times-trig integral solvable by parts.

    Step-by-Step Solution

    1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
    2. $\int\cos\sqrt x\,dx = \int\cos t\cdot 2t\,dt = 2\int t\cos t\,dt$.
    3. Integrate by parts: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$. …

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11. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

    ›Reveal solutionSolution

    Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

    Concept and Intuition

    When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

    Step-by-Step Solution

    1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
    2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
    3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
    4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$. …

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12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{2\cos x + 3\sin x + 4} = \dfrac{2}{\sqrt{3}}f(x) + c$, then $f\left(\dfrac{2\pi}{3}\right) =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{8}$ (C) $\dfrac{5\pi}{12}$ (D) $\dfrac{5\pi}{8}$

    ›Reveal solutionSolution

    This is a Weierstrass ($t=\tan(x/2)$) substitution problem for a linear combination of sine and cosine plus a constant in the denominator. Evaluating $f$ at the given point gives $\dfrac{5\pi}{12}$, option (C).

    Concept and Intuition

    Whenever the denominator mixes $\sin x$, $\cos x$ and a constant, the universal substitution $t=\tan(x/2)$ (with $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2\,dt}{1+t^2}$) converts the trigonometric denominator into a plain quadratic in $t$, reducing the whole problem to a standard $\int\frac{dt}{\text{quadratic}}$ that integrates to an arctangent.

    Step-by-Step Solution

    1. Substitute: denominator becomes

    $2\cdot\frac{1-t^2}{1+t^2}+3\cdot\frac{2t}{1+t^2}+4=\frac{2-2t^2+6t+4+4t^2}{1+t^2}=\frac{2t^2+6t+6}{1+t^2}.$

    2. The integral becomes $\displaystyle\int\frac{2\,dt/(1+t^2)}{(2t^2+6t+6)/(1+t^2)}=\int\frac{2\,dt}{2t^2+6t+6}=\int\frac{dt}{t^2+3t+3}$.
    3. Complete the square: $t^2+3t+3=\left(t+\tfrac32\right)^2+\tfrac34$, so $\int\frac{dt}{\left(t+\frac32\right)^2+\frac34}=\frac{1}{\sqrt{3}/2}\arctan\!\left(\frac{t+3/2}{\sqrt3/2}\right)+c=\frac{2}{\sqrt3}\arctan\!\left(\frac{2t+3}{\sqrt3}\right)+c.$ …

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