Q.Evaluate the definite integral: $\int_{0}^{2} \frac{6x+3}{x^2+4} \, dx$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\displaystyle\int_{\log 2}^{\log 3} \dfrac{e^{3x} - 3e^x}{e^{2x} - 1}\,dx =$ (A) $\log\left(\dfrac{2e}{3}\right)$ (B) $\log\left(\dfrac{4e}{3}\right)$ (C) $\dfrac{2e}{3}$ (D) $\dfrac{4e}{3}$

   ›Reveal solutionSolution

   A substitution $t=e^x$ turns this exponential integral into a rational one; the value works out to $\log(2e/3)$.

   Concept and Intuition

   Whenever an integral is built entirely from powers of $e^x$, substituting $t=e^x$ converts it into an algebraic (rational function) integral, which is usually far easier to handle with partial fractions.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^xdx=t\,dx$, i.e. $dx=dt/t$. When $x=\log2$, $t=2$; when $x=\log3$, $t=3$.
   2. Rewrite the integrand: $\dfrac{e^{3x}-3e^x}{e^{2x}-1}=\dfrac{t^3-3t}{t^2-1}$. Multiplying by $dx=dt/t$ gives $\dfrac{t^3-3t}{t(t^2-1)}dt=\dfrac{t^2-3}{t^2-1}dt$.
   3. So the integral becomes $\displaystyle\int_2^3\dfrac{t^2-3}{t^2-1}dt$.
   4. Split: $\dfrac{t^2-3}{t^2-1}=\dfrac{(t^2-1)-2}{t^2-1}=1-\dfrac{2}{t^2-1}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

   ›Reveal solutionSolution

   The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

   Concept and Intuition

   Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

   Step-by-Step Solution

   1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
   2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
   3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
   4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
   5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $\int\frac{dt}{\left(t+\frac32\right)^2+\frac34}=\frac{1}{\sqrt{3}/2}\arctan\!\left(\frac{t+3/2}{\sqrt3/2}\right)+c=\frac{2}{\sqrt3}\arctan\!\left(\frac{2t+3}{\sqrt3}\right)+c.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int_{\log4}^{\log5} \dfrac{e^{2x}+e^x}{e^{2x}-5e^x+6}\,dx=$ (A) $\log\left(\dfrac{64}{9}\right)$ (B) $\log\left(\dfrac{256}{81}\right)$ (C) $\log\left(\dfrac{32}{3}\right)$ (D) $\log\left(\dfrac{128}{27}\right)$

   ›Reveal solutionSolution

   Substituting $t=e^x$ turns the integral into a simple rational-function integral that evaluates to $\log(128/27)$.

   Concept and Intuition

   Whenever an integrand is built entirely out of $e^x$ (here $e^{2x}$ and $e^x$), the substitution $t=e^x$ converts it into a rational function of $t$, which can then be handled by partial fractions — a standard technique for turning transcendental-looking integrals into algebraic ones.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^x\,dx=t\,dx$, i.e. $dx=\dfrac{dt}{t}$. When $x=\log 4$, $t=4$; when $x=\log 5$, $t=5$.
   2. Rewrite the integrand: $e^{2x}+e^x=t^2+t=t(t+1)$, and $e^{2x}-5e^x+6=t^2-5t+6=(t-2)(t-3)$.
   3. So $\displaystyle\int \frac{t(t+1)}{(t-2)(t-3)}\cdot\frac{dt}{t}=\int_4^5 \frac{t+1}{(t-2)(t-3)}\,dt$.
   4. Partial fractions: $\dfrac{t+1}{(t-2)(t-3)}=\dfrac{A}{t-2}+\dfrac{B}{t-3}$. At $t=2$: $A=\dfrac{3}{-1}=-3$. At $t=3$: $B=\dfrac{4}{1}=4$.
   5. So the integral is $\displaystyle\int_4^5\left(\frac{-3}{t-2}+\frac{4}{t-3}\right)dt=\Big[-3\log|t-2|+4\log|t-3|\Big]_4^5$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\displaystyle\int (x^6 + x^4 + x^2)\sqrt{2x^4 + 3x^2 + 6}\,dx = f(x) + c$, then $f(3) =$ (A) $\dfrac{3}{2}(95)^{3/2}$ (B) $\dfrac{3}{2}(195)^{3/2}$ (C) $\dfrac{3}{2}(265)^{3/2}$ (D) $\dfrac{3}{2}(175)^{3/2}$

   ›Reveal solutionSolution

   Guess the antiderivative in the form $(ax^3+bx)(2x^4+3x^2+6)^{3/2}$, match

   coefficients by differentiating, solve for $a,b$, then evaluate at $x=3$ to get

   $f(3)=\dfrac32(195)^{3/2}$.

   Concept and Intuition

   When an integrand is a polynomial times the square root of another polynomial,

   and the polynomial factor's degree matches what you'd get by differentiating a

   $(\text{poly})\cdot(\text{radicand})^{3/2}$ ansatz, the fastest route is the

   "reverse chain rule with an undetermined polynomial coefficient" method: guess

   the shape of the antiderivative with unknown coefficients, differentiate, and

   match coefficients of like powers of $x$ to solve for them.

   Step-by-Step Solution

   1. Guess $f(x)=(ax^3+bx)\,(2x^4+3x^2+6)^{3/2}$.
   2. Differentiate (product + chain rule), then factor out $(2x^4+3x^2+6)^{1/2}$: $f'(x)=(2x^4+3x^2+6)^{1/2}\Big[(3ax^2+b)(2x^4+3x^2+6)+(ax^3+bx)(12x^3+9x)\Big]$.
   3. Expand the bracket: it collects to $18a\,x^6+(18a+14b)x^4+(18a+12b)x^2+6b$.
   4. Match this to the target $x^6+x^4+x^2$ (constant term $0$): $6b=0\Rightarrow b=0$; then $18a=1\Rightarrow a=\dfrac1{18}$.
   5. Check remaining coefficients with $a=1/18,\ b=0$: $18a+14b=1$ ✓ (matches $x^4$ coefficient $1$), $18a+12b=1$ ✓ (matches $x^2$ coefficient $1$). All consistent.
   6. So $f(x)=\dfrac{x^3}{18}(2x^4+3x^2+6)^{3/2}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\sin 2\theta}{(1 + \sin\theta)(3 - \cos 2\theta)}\, d\theta = \dfrac{1}{2}\tan^{-1}(\sin\theta) + \dfrac{1}{4}\log(f(\theta)) + c$ then $f\left(\dfrac{\pi}{2}\right) - f(0) =$ (A) $\dfrac{1}{2}$ (B) $-\dfrac{1}{2}$ (C) $0$ (D) $-\dfrac{3}{4}$

   ›Reveal solutionSolution

   Reducing the integral via $t=\sin\theta$ and partial fractions identifies $f(\theta)=\dfrac{1+\sin^2\theta}{(1+\sin\theta)^2}$, giving $f(\pi/2)-f(0)=-1/2$.

   Concept and Intuition

   Double-angle identities collapse $\sin2\theta$ and $3-\cos2\theta$ into expressions purely in $\sin\theta$ and $\cos\theta$; then $t=\sin\theta$ (since $\cos\theta\,d\theta = dt$ appears naturally) turns the whole thing into a rational-function integral solvable by partial fractions — a very standard pattern for trig integrals with even powers/mixed degree-2 denominators.

   Step-by-Step Solution

   1. $\sin2\theta = 2\sin\theta\cos\theta$; $\cos2\theta = 1-2\sin^2\theta \Rightarrow 3-\cos2\theta = 2+2\sin^2\theta = 2(1+\sin^2\theta)$.
   2. Integrand becomes $\dfrac{2\sin\theta\cos\theta}{(1+\sin\theta)\cdot 2(1+\sin^2\theta)} = \dfrac{\sin\theta\cos\theta}{(1+\sin\theta)(1+\sin^2\theta)}$.
   3. Substitute $t=\sin\theta$, $dt=\cos\theta\,d\theta$: integral $= \displaystyle\int \dfrac{t}{(1+t)(1+t^2)}dt$.
   4. Partial fractions: $\dfrac{t}{(1+t)(1+t^2)} = \dfrac{-1/2}{1+t} + \dfrac{(1/2)t+1/2}{1+t^2}$ (solve $t=A(1+t^2)+(Bt+C)(1+t)$, giving $A=-1/2,\ B=1/2,\ C=1/2$).
   5. Integrate: $-\dfrac12\log(1+t) + \dfrac14\log(1+t^2) + \dfrac12\tan^{-1}t + c$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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 $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $x^{-5}dx = -\dfrac{dt}{4}$. …

    🔒 Start your 14-day free trial to see the full solution →