Q.Integrate the following function: $\frac{x^2}{(2+3x^3)^3}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^3}{\sqrt{1+x^2}}\,dx = A(1+x^2)^{3/2}+B(1+x^2)^{1/2}+C$, then $A+B=$ (A) $2/3$ (B) $-2/3$ (C) $1/3$ (D) $-1/3$

   ›Reveal solutionSolution

   The substitution $u=1+x^2$ turns the integral into a simple power-rule computation, giving $A=1/3$ and $B=-1$, so $A+B=-2/3$.

   Concept and Intuition

   Whenever the integrand has an odd power of $x$ alongside a function of $x^2$ (here $\sqrt{1+x^2}$), substituting $u=1+x^2$ (so $du=2x\,dx$) converts the odd-power part into a polynomial in $u$, making the integral elementary.

   Step-by-Step Solution

   1. Let $u=1+x^2$, $du=2x\,dx$, and $x^2=u-1$.
   2. $x^3\,dx=x^2\cdot x\,dx=(u-1)\cdot\dfrac{du}{2}$.
   3. $\displaystyle\int\frac{x^3}{\sqrt{1+x^2}}dx=\int\frac{(u-1)}{\sqrt u}\cdot\frac{du}{2}=\frac12\int\left(u^{1/2}-u^{-1/2}\right)du$.
   4. $=\frac12\left(\dfrac{2}{3}u^{3/2}-2u^{1/2}\right)+C=\dfrac13u^{3/2}-u^{1/2}+C$. …

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2. 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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3. 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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4. 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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5. 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: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$. …

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6. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.$\displaystyle\int_{2/e}^{1/e} \dfrac{1}{x(\log x)^{1/3}}\,dx =$ (A) $\dfrac{3}{2}\{1 + (\log(2) - 1)^{2/3}\}$ (B) $1$ (C) $\dfrac{3}{2}\{1 + (\log(2) + 1)^{3/2}\}$ (D) $\dfrac{3}{2}\{1 - (\log(2) - 1)^{2/3}\}$

   ›Reveal solutionSolution

   Substitute $u=\log x$ to turn the integral into $\int u^{-1/3}du$, then evaluate between the transformed limits $u=\log2-1$ and $u=-1$.

   Concept and Intuition

   An integrand of the form $\dfrac{1}{x\cdot g(\log x)}$ always calls for the substitution $u=\log x$, since $du = dx/x$ removes the $x$ and $1/x$ entirely, leaving a pure power of $u$.

   Step-by-Step Solution

   1. Let $u=\log x$, so $du = dx/x$.
   2. Limits: at $x=2/e$, $u=\log(2/e)=\log2-1$. At $x=1/e$, $u=\log(1/e)=-1$.
   3. The integral becomes $\displaystyle\int_{\log2-1}^{-1} u^{-1/3}\,du = \left[\frac{u^{2/3}}{2/3}\right]_{\log2-1}^{-1} = \frac32\Big[u^{2/3}\Big]_{\log2-1}^{-1}$.
   4. Using the real cube root, $(-1)^{2/3} = \big((-1)^{1/3}\big)^2 = (-1)^2 = 1$.
   5. So the value is $\dfrac32\big[1-(\log2-1)^{2/3}\big]$.

   Common Mistakes …

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7. 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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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$ (A) $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$ (B) $\dfrac{2}{(1+\sqrt{x})^{2022}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{\sqrt{x}}{2021}\right] + C$ (C) $\dfrac{2}{(1+\sqrt{x})}\left[\dfrac{(1+\sqrt{x})^{2022}}{2022} - \dfrac{(1+\sqrt{x})^{2021}}{2021}\right] + C$ (D) $\dfrac{1}{(1+\sqrt{x})^2}\left[\dfrac{1}{(1+\sqrt{x})^{1010}} - \dfrac{1}{(1+\sqrt{x})^{1011}}\right] + C$

   ›Reveal solutionSolution

   Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$; back-substituting and factoring reproduces option (A)'s bracketed form.

   Concept and Intuition

   Whenever an integrand is a function purely of $1+\sqrt x$, the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$, $x=(t-1)^2$) turns the messy radical expression into a clean power of $t$, and the pieces of $dx$ that are left over ($2(t-1)\,dt$) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

   Step-by-Step Solution

   1. Let $t=1+\sqrt x$. Then $\sqrt x=t-1$, $x=(t-1)^2$, and $dx=2(t-1)\,dt$.
   2. The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$.
   3. Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$, and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$, combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$. …

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9. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.If $\displaystyle\int \sqrt[3]{x}\left\{1 + \sqrt[3]{x^4}\right\}^{1/7} dx = A\left(1 + \sqrt[3]{x^4}\right)^B + C$, then value of $AB =$ ____ (A) $\dfrac{3}{2}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{3}{32}$ (D) $\dfrac{4}{3}$

   ›Reveal solutionSolution

   A direct substitution $u=1+x^{4/3}$ turns the integral into a simple power rule, from which $A$ and $B$ are read off and multiplied.

   Concept and Intuition

   When the integrand contains $x^{1/3}$ times a function of $x^{4/3}$, substituting $u=1+x^{4/3}$ (whose derivative involves exactly $x^{1/3}dx$) is the natural simplification.

   Step-by-Step Solution

   1. Let $u=1+x^{4/3}$. Then $du=\dfrac43x^{1/3}dx\Rightarrow x^{1/3}dx=\dfrac34du$.
   2. The integral $\displaystyle\int x^{1/3}(1+x^{4/3})^{1/7}dx=\dfrac34\int u^{1/7}du$.
   3. $\displaystyle\int u^{1/7}du=\dfrac{u^{8/7}}{8/7}=\dfrac78u^{8/7}$.
   4. So the integral $=\dfrac34\cdot\dfrac78u^{8/7}+C=\dfrac{21}{32}\left(1+x^{4/3}\right)^{8/7}+C$. …

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10. 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 $f(x)=\dfrac{x^3}{18}(2x^4+3x^2+6)^{3/2}$. …

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11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$0 < x < 1$, $\int \dfrac{dx}{\sqrt{x^2 - x^5}} = \dfrac{1}{3}\log|f(x)| + C$, then $f(1/2) =$ (A) $\dfrac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}}$ (B) $\dfrac{\sqrt{8} + \sqrt{7}}{\sqrt{8} - \sqrt{7}}$ (C) $2(\sqrt{8} - \sqrt{7})$ (D) $2(\sqrt{8} - \sqrt{7})^2$

    ›Reveal solutionSolution

    Substituting $t=x^3$ then $1-t=w^2$ integrates $\int dx/(x\sqrt{1-x^3})$ cleanly to $\tfrac13\log\left|\tfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|$, and evaluating at $x=1/2$ gives $\dfrac{\sqrt8-\sqrt7}{\sqrt8+\sqrt7}$.

    Concept and Intuition

    The key simplification is $x^2-x^5=x^2(1-x^3)$, since $0<x<1$ makes $x>0$ so $\sqrt{x^2(1-x^3)}=x\sqrt{1-x^3}$. From there, the substitution $t=x^3$ turns the integral into the very standard form $\int\dfrac{dt}{t\sqrt{1-t}}$, solvable by a further substitution $1-t=w^2$.

    Step-by-Step Solution

    1. Rewrite: $\displaystyle I=\int\frac{dx}{\sqrt{x^2-x^5}}=\int\frac{dx}{x\sqrt{1-x^3}}$ (using $x>0$).
    2. Let $t=x^3\Rightarrow dt=3x^2dx\Rightarrow dx=\dfrac{dt}{3x^2}$. Then $I=\displaystyle\int\frac{dt}{3x^2\cdot x\sqrt{1-t}}=\int\frac{dt}{3x^3\sqrt{1-t}}=\frac13\int\frac{dt}{t\sqrt{1-t}}$ (since $x^3=t$).
    3. Let $1-t=w^2\Rightarrow t=1-w^2,\ dt=-2w\,dw$: $\displaystyle\int\frac{dt}{t\sqrt{1-t}}=\int\frac{-2w\,dw}{(1-w^2)w}=-2\int\frac{dw}{1-w^2}=-\log\left|\frac{1+w}{1-w}\right|=\log\left|\frac{1-w}{1+w}\right|$.
    4. So $I=\dfrac13\log\left|\dfrac{1-w}{1+w}\right|+C$ where $w=\sqrt{1-t}=\sqrt{1-x^3}$, i.e. $f(x)=\dfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.
    5. Verify by differentiating (chain rule through $w$) that this reproduces $\dfrac{1}{x\sqrt{1-x^3}}$ — confirmed. …

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12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If $f(x) = \dfrac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$, then $\displaystyle\int x^{n-2} f(x)\, dx =$ (A) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (B) $\dfrac{1}{n-1} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (C) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 + \frac{1}{n}} + C$ (D) $\dfrac{1}{n+1} (1 + nx^n)^{1 + \frac{1}{n}} + C$

    ›Reveal solutionSolution

    A direct substitution $u=1+nx^n$ turns the integral into a simple power rule.

    Concept and Intuition

    The integrand's power of $x$ (namely $x^{n-1}$) is exactly proportional to the derivative of $u=1+nx^n$, which is the classic signal to substitute.

    Step-by-Step Solution

    1. $x^{n-2}f(x)=x^{n-2}\cdot\dfrac{x}{(1+nx^n)^{1/n}}=\dfrac{x^{n-1}}{(1+nx^n)^{1/n}}$.
    2. Let $u=1+nx^n$. Then $du=n\cdot n x^{n-1}dx=n^2x^{n-1}dx$, so $x^{n-1}dx=\dfrac{du}{n^2}$.
    3. Integral becomes $\displaystyle\int u^{-1/n}\cdot\dfrac{du}{n^2}=\dfrac{1}{n^2}\cdot\dfrac{u^{1-1/n}}{1-1/n}+C$.
    4. Simplify: $\dfrac{1}{n^2}\cdot\dfrac{1}{\frac{n-1}{n}}=\dfrac{1}{n^2}\cdot\dfrac{n}{n-1}=\dfrac{1}{n(n-1)}$. …

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