Q.$\int \dfrac{x^9}{(4x^2+1)^6}\,dx$ is equal to
(A) $\dfrac{1}{5x}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(B) $\dfrac{1}{5}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(C) $\dfrac{1}{10x}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(D) $\dfrac{1}{10}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x^5\sqrt[5]{x^5+1}}\,dx=$ (A) $\dfrac{4}{\sqrt[5]{x^5+1}}+c$ (B) $4x^4(x^5+1)^{4/5}+c$ (C) $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$ (D) $-\dfrac{(x^5+1)^{4/5}}{4x^5}+c$

   ›Reveal solutionSolution

   Rewriting the integrand to expose $1+x^{-5}$ as the natural substitution variable solves this cleanly; the answer is $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Concept and Intuition

   When an integral mixes a power of $x$ with a root of a polynomial in $x$, factoring out the highest power of $x$ from inside the root often converts the expression into a function of $1/x$ (or $x^{-5}$ here), whose derivative is already present elsewhere in the integrand — a clean substitution.

   Step-by-Step Solution

   1. $(x^5+1)^{-1/5}=\left(x^5\left(1+x^{-5}\right)\right)^{-1/5}=x^{-1}(1+x^{-5})^{-1/5}$.
   2. So the integrand $x^{-5}(x^5+1)^{-1/5}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $t=1+x^{-5}$, so $dt=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dt}{5}$.
   4. Integral $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$. …

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2. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x-3)^{4/5}(x+1)^{6/5}} =$ (A) $\dfrac{5}{4}\sqrt[5]{\dfrac{x-3}{x+1}} + C$ (B) $\dfrac{5}{4}\left(\dfrac{x+1}{x-3}\right)^{1/5} + C$ (C) $\dfrac{1}{5}\left(\dfrac{x-3}{x+1}\right)^{1/5} + C$ (D) $\dfrac{5}{4}\left(\dfrac{x-3}{x+4}\right)^{4/5} + C$

   ›Reveal solutionSolution

   Splitting off $(x+1)^2$ turns the integrand into a pure power of $t=\frac{x-3}{x+1}$, giving $\frac54\left(\frac{x-3}{x+1}\right)^{1/5}+C$.

   Concept and Intuition

   When an integrand has the form $(x-a)^{p}(x-b)^{q}$ with $p+q$ an integer (here $\frac45+\frac65=2$), factoring out $(x-b)^{p+q}$ and substituting $t=\frac{x-a}{x-b}$ collapses the whole thing to a simple power of $t$ — a standard trick for these "unequal fractional exponent" integrals.

   Step-by-Step Solution

   1. $(x-3)^{4/5}(x+1)^{6/5} = (x+1)^2\left[\dfrac{x-3}{x+1}\right]^{4/5}$ (factoring out $(x+1)^{4/5+6/5}=(x+1)^2$).
   2. So the integrand is $(x+1)^{-2}\left[\dfrac{x-3}{x+1}\right]^{-4/5}$.
   3. Let $t=\dfrac{x-3}{x+1}$. Then $\dfrac{dt}{dx} = \dfrac{(x+1)-(x-3)}{(x+1)^2} = \dfrac{4}{(x+1)^2}$, so $(x+1)^{-2}dx = \dfrac{dt}{4}$.
   4. The integral becomes $\dfrac14\displaystyle\int t^{-4/5}\,dt = \dfrac14\cdot\dfrac{t^{1/5}}{1/5}+C = \dfrac54\,t^{1/5}+C$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x^5+1)^{6/5}} =$ (A) $\dfrac{1}{\sqrt[5]{x^5+1}} + c$ (B) $\dfrac{\sqrt[5]{x^5+1}}{x} + c$ (C) $\dfrac{x}{\sqrt[5]{x^5+1}} + c$ (D) $\sqrt[5]{x^5+1} + c$

   ›Reveal solutionSolution

   Recognizing the integrand as the derivative of $\dfrac{x}{\sqrt[5]{x^5+1}}$ (verified by direct differentiation) gives the antiderivative immediately.

   Concept and Intuition

   For integrals of the form $\int\frac{dx}{(x^n+1)^{(n+1)/n}}$, a useful trick is to guess that the antiderivative looks like $\frac{x}{(x^n+1)^{1/n}}$ (a ratio designed to make the product-rule differentiation collapse nicely), and then verify by differentiating it — if it reproduces the integrand exactly, we're done. This is often faster than a substitution for this particular family.

   Step-by-Step Solution

   1. Guess the antiderivative $g(x) = \dfrac{x}{(x^5+1)^{1/5}} = x(x^5+1)^{-1/5}$.
   2. Differentiate using the product rule: $g'(x) = (x^5+1)^{-1/5} + x\cdot\left(-\dfrac15\right)(x^5+1)^{-6/5}\cdot5x^4$.
   3. Simplify the second term: $x\cdot\left(-\dfrac15\right)(5x^4)(x^5+1)^{-6/5} = -x^5(x^5+1)^{-6/5}$.
   4. So $g'(x) = (x^5+1)^{-1/5} - x^5(x^5+1)^{-6/5}$.
   5. Factor out $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$. …

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4. 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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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{4+5\cos x} =$ (A) $-\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (B) $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (C) $-\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$ (D) $\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$

   ›Reveal solutionSolution

   The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

   Concept and Intuition

   For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$, $dx=\tfrac{2\,dt}{1+t^2}$) always converts the integral into a rational function of $t$ alone.

   Step-by-Step Solution

   1. $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, with $t=\tan(x/2)$.
   2. $4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$.
   3. Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$.
   4. Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$. …

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6. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^4 x}{\left(\sin^2 x + \sin^{-3} x \cos^5 x\right)^{3}}\;dx =$ (A) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-2} + C$ (B) $\dfrac{1}{10}\left(1 + \cot^2 x\right)^{-5} + C$ (C) $\dfrac{1}{10}\left(1 + \cot^5 x\right)^{-2} + C$ (D) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-5} + C$

   ›Reveal solutionSolution

   Factoring $\sin^2x$ out of the denominator turns it into $\sin^2x(1+\cot^5x)$, and the substitution $t=1+\cot^5x$ makes the whole integral a simple power-rule integration, giving $\dfrac{1}{10}(1+\cot^5x)^{-2}+C$.

   Concept and Intuition

   Integrals with mixed powers of $\sin x$ and $\cos x$ in odd/negative combinations often simplify beautifully once you factor out a common power to expose a $(1+\cot^n x)$ or $(1+\tan^n x)$ structure — this is exactly the kind of expression whose derivative (via chain rule) reproduces $\cot^{n-1}x\csc^2x$, matching what's left over in the integrand.

   Step-by-Step Solution

   1. Denominator: $\sin^2x+\sin^{-3}x\cos^5x$. Factor out $\sin^2x$: $=\sin^2x\left[1+\dfrac{\cos^5x}{\sin^5x}\right]=\sin^2x\,(1+\cot^5x)$.
   2. So the full denominator cubed: $\left[\sin^2x(1+\cot^5x)\right]^3=\sin^6x\,(1+\cot^5x)^3$.
   3. Integrand: $\dfrac{\cos^4x}{\sin^6x(1+\cot^5x)^3} = \dfrac{\cos^4x}{\sin^4x}\cdot\dfrac{1}{\sin^2x}\cdot(1+\cot^5x)^{-3}=\cot^4x\,\csc^2x\,(1+\cot^5x)^{-3}$.
   4. Substitute $t=1+\cot^5x$. Then $\dfrac{dt}{dx}=5\cot^4x\cdot(-\csc^2x)=-5\cot^4x\csc^2x$, so $\cot^4x\csc^2x\,dx=-\dfrac{dt}{5}$. …

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7. 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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8. 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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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\int x^5 e^{-4x^3}dx = \frac{1}{48}e^{-4x^3}f(x) + c$, then $f(x) =$ (A) $-2x^3 - 1$ (B) $-4x^3 - 1$ (C) $-2x^2 + 1$ (D) $4x^3 + 1$

   ›Reveal solutionSolution

   Substituting $u=x^3$ converts the integral into a simple integration-by-parts problem $\int ue^{-4u}du$; matching the result to the given form yields $f(x)=-4x^3-1$.

   Concept and Intuition

   The presence of $x^5$ alongside $e^{-4x^3}$ is a strong hint to substitute $u=x^3$, since then $x^2\,dx$ (part of $du$) combines with the remaining $x^3=u$ to leave a clean polynomial-times-exponential integral, solvable by the standard integration-by-parts reduction formula for $\int u\,e^{ku}du$.

   Step-by-Step Solution

   1. Let $u=x^3$, so $du=3x^2\,dx \Rightarrow x^2dx=\dfrac{du}{3}$.
   2. Rewrite $x^5e^{-4x^3}dx = x^3\cdot x^2 e^{-4x^3}dx = u\,e^{-4u}\cdot\dfrac{du}{3}$.
   3. So the integral becomes $\dfrac13\displaystyle\int u\,e^{-4u}\,du$.
   4. Integrate by parts with first function $u$, second $e^{-4u}$: $\int u e^{-4u}du = u\cdot\left(-\dfrac{e^{-4u}}{4}\right) - \int\left(-\dfrac{e^{-4u}}{4}\right)du = -\dfrac{u}{4}e^{-4u} - \dfrac{1}{16}e^{-4u}$. …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{x^3 \tan^{-1} x^4}{1 + x^8}\,dx =$ (A) $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$ (B) $\dfrac{(\tan^{-1}(x^4))^3}{3} + c$ (C) $\dfrac{(\tan^{-1}(x^4))^2}{4} + c$ (D) $\dfrac{(\tan^{-1}(x^4))^2}{2} + c$

    ›Reveal solutionSolution

    A double substitution (first $u=x^4$, then $v=\tan^{-1}u$) turns this into a trivial $\int v\,dv$. Answer: $\dfrac{(\tan^{-1}x^4)^2}{8}+c$.

    Concept and Intuition

    The presence of $x^3\,dx$ alongside $x^4$ inside the arctan and $x^8=(x^4)^2$ in the denominator is a strong signal to substitute $u=x^4$ first. After that, the structure $\tan^{-1}u \cdot \dfrac{du}{1+u^2}$ is exactly of the form "function times its own derivative," solved by a second substitution.

    Step-by-Step Solution

    1. Let $u=x^4$, so $du=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{du}{4}$.
    2. The integral becomes $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$. …

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11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.$\int(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{\frac{1}{m}}dx=$ (A) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m+1}{m}}+C$ (B) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$ (C) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6\right)^{\frac{m+1}{m}}+C$ (D) $\dfrac{1}{6(m-1)}\left(2x^{3m}+mx^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$

    ›Reveal solutionSolution

    Recognising that $2x^{3m}+3x^{2m}+6x^m$ equals $x^m$ times the bracket under the $1/m$-power root lets the whole integrand be rewritten as (a constant times) $g^{1/m}g'$ for $g=2x^{3m}+3x^{2m}+6x^m$ — a pure "power rule" integral.

    Concept and Intuition

    Whenever an integrand looks like $g(x)^{1/m}\cdot g'(x)$ (up to a constant factor), the antiderivative is immediately $\dfrac{g^{1/m+1}}{1/m+1}$. The main work here is algebraic: spotting that the "outside" factor $(x^{3m}+x^{2m}+x^m)$ is secretly related to the derivative of $g=2x^{3m}+3x^{2m}+6x^m$, and that the "inside" bracket $(2x^{2m}+3x^m+6)$ is just $g/x^m$.

    Step-by-Step Solution

    1. Let $g=2x^{3m}+3x^{2m}+6x^m$. Factor: $g=x^m(2x^{2m}+3x^m+6)$, so $2x^{2m}+3x^m+6=g/x^m$, and $(2x^{2m}+3x^m+6)^{1/m}=g^{1/m}/x$ (since $(g/x^m)^{1/m}=g^{1/m}/x$).
    2. Differentiate $g$: $g'=6mx^{3m-1}+6mx^{2m-1}+6mx^{m-1}=6mx^{m-1}(x^{2m}+x^m+1)$.
    3. Rewrite the integrand: $(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m}=x^m(x^{2m}+x^m+1)\cdot\dfrac{g^{1/m}}x=x^{m-1}(x^{2m}+x^m+1)\,g^{1/m}$.
    4. From step 2, $x^{m-1}(x^{2m}+x^m+1)=\dfrac{g'}{6m}$. So the integrand equals $\dfrac{g'}{6m}g^{1/m}$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{\sqrt{1 - x^4}}{x^7}\, dx = f(x)\left\{\sqrt{1 - x^4}\right\}^n + c$, then $(f(x))^n =$ ______ (A) $\dfrac{-1}{6x^6}$ (B) $\dfrac{-1}{216x^{18}}$ (C) $\dfrac{1}{36x^{12}}$ (D) $\dfrac{1}{216x^{18}}$

    ›Reveal solutionSolution

    A reduction-formula-style substitution $t=x^{-2}$ turns the integral into a simple power form; matching to the given answer form identifies $f(x)$ and $n$, then $(f(x))^n$ follows directly.

    Concept and Intuition

    Integrals of the form $\int x^m(a+bx^n)^p\,dx$ can often be simplified by substituting $t=x^{-n}$ when $(m+1)/n$ isn't an integer but $(m+1)/n+p$ is — exactly the situation here with $m=-7$, $n=4$, $p=1/2$.

    Step-by-Step Solution

    1. Write the integral as $\displaystyle\int x^{-7}(1-x^4)^{1/2}\,dx$.
    2. Substitute $t = x^{-2}$, so $dt = -2x^{-3}dx \Rightarrow dx = -\dfrac{x^3}{2}dt$, and $x^{-4}=t^2$.
    3. $x^{-7}dx = x^{-7}\cdot\left(-\dfrac{x^3}{2}\right)dt = -\dfrac{x^{-4}}{2}dt = -\dfrac{t^2}{2}dt$.
    4. $1-x^4 = 1-\dfrac{1}{t^2} = \dfrac{t^2-1}{t^2}$, so $\sqrt{1-x^4}=\dfrac{\sqrt{t^2-1}}{t}$ (for $t>0$).
    5. The integral becomes $\displaystyle\int \dfrac{\sqrt{t^2-1}}{t}\cdot\left(-\dfrac{t^2}{2}\right)dt = -\dfrac12\int t\sqrt{t^2-1}\,dt$.
    6. Let $s=t^2-1$, $ds=2t\,dt$: $-\dfrac12\int \sqrt{s}\cdot\dfrac{ds}{2} = -\dfrac14\cdot\dfrac{2}{3}s^{3/2} = -\dfrac16 (t^2-1)^{3/2}+c$. …

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