Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Indefinite Integrals question

2021 · 31 Aug · Shift 1 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Indefinite Integrals
  5. /2021 · 31 Aug · Shift 1 · Q28

Indefinite Integrals question

2021 · 31 Aug · Shift 1 · Q28

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
The integral ∫1(x−1)3(x+2)54 dx\int {{1 \over {\sqrt[4]{{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx∫4(x−1)3(x+2)5​1​dx is equal to : (where C is a constant of integration)
  1. A
    34(x+2x−1)14+C{3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{1 \over 4}}} + C43​(x−1x+2​)41​+C
  2. B
    34(x+2x−1)54+C{3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{5 \over 4}}} + C43​(x−1x+2​)45​+C
  3. C
    43(x−1x+2)14+C{4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C34​(x+2x−1​)41​+C
  4. D
    43(x−1x+2)54+C{4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{5 \over 4}}} + C34​(x+2x−1​)45​+C
View written solutionFree

Correct answer: C

  1. Rewrite the integrand

Given

I=∫1(x−1)3(x+2)54 dxI=\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dxI=∫4(x−1)3(x+2)5​1​dx

we write it using exponents:

I=∫(x−1)−3/4(x+2)−5/4 dxI=\int (x-1)^{-3/4}(x+2)^{-5/4}\,dxI=∫(x−1)−3/4(x+2)−5/4dx
  1. Look for a useful derivative form

Notice the options involve expressions like

(x−1x+2)1/4\left(\frac{x-1}{x+2}\right)^{1/4}(x+2x−1​)1/4

So let

f(x)=(x−1x+2)1/4f(x)=\left(\frac{x-1}{x+2}\right)^{1/4}f(x)=(x+2x−1​)1/4

Then

f(x)=(x−1x+2)1/4f(x)=\left(\frac{x-1}{x+2}\right)^{1/4}f(x)=(x+2x−1​)1/4

Differentiate using chain rule:

f′(x)=14(x−1x+2)−3/4⋅ddx(x−1x+2)f'(x)=\frac14\left(\frac{x-1}{x+2}\right)^{-3/4}\cdot \frac{d}{dx}\left(\frac{x-1}{x+2}\right)f′(x)=41​(x+2x−1​)−3/4⋅dxd​(x+2x−1​)

Now,

ddx(x−1x+2)=(x+2)−(x−1)(x+2)2=3(x+2)2\frac{d}{dx}\left(\frac{x-1}{x+2}\right)=\frac{(x+2)-(x-1)}{(x+2)^2}=\frac{3}{(x+2)^2}dxd​(x+2x−1​)=(x+2)2(x+2)−(x−1)​=(x+2)23​

Hence,

f′(x)=14(x−1x+2)−3/4⋅3(x+2)2f'(x)=\frac14\left(\frac{x-1}{x+2}\right)^{-3/4}\cdot \frac{3}{(x+2)^2}f′(x)=41​(x+2x−1​)−3/4⋅(x+2)23​
  1. Simplify

Since

(x−1x+2)−3/4=(x+2x−1)3/4\left(\frac{x-1}{x+2}\right)^{-3/4}=\left(\frac{x+2}{x-1}\right)^{3/4}(x+2x−1​)−3/4=(x−1x+2​)3/4

we get

f′(x)=34⋅1(x+2)2⋅(x+2x−1)3/4f'(x)=\frac34\cdot \frac{1}{(x+2)^2}\cdot \left(\frac{x+2}{x-1}\right)^{3/4}f′(x)=43​⋅(x+2)21​⋅(x−1x+2​)3/4 =34 (x+2)−2(x+2)3/4(x−1)−3/4=\frac34\,(x+2)^{-2}(x+2)^{3/4}(x-1)^{-3/4}=43​(x+2)−2(x+2)3/4(x−1)−3/4 =34 (x−1)−3/4(x+2)−5/4=\frac34\,(x-1)^{-3/4}(x+2)^{-5/4}=43​(x−1)−3/4(x+2)−5/4

This is exactly

f′(x)=34⋅1(x−1)3(x+2)54f'(x)=\frac34\cdot \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}f′(x)=43​⋅4(x−1)3(x+2)5​1​

Therefore,

1(x−1)3(x+2)54=43f′(x)\frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}=\frac43 f'(x)4(x−1)3(x+2)5​1​=34​f′(x)
  1. Integrate

So,

I=∫1(x−1)3(x+2)54 dx=43f(x)+CI=\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dx =\frac43 f(x)+CI=∫4(x−1)3(x+2)5​1​dx=34​f(x)+C I=43(x−1x+2)1/4+CI=\frac43\left(\frac{x-1}{x+2}\right)^{1/4}+CI=34​(x+2x−1​)1/4+C
  1. Match with options

This is exactly Option C:

43(x−1x+2)1/4+C\boxed{\frac43\left(\frac{x-1}{x+2}\right)^{1/4}+C}34​(x+2x−1​)1/4+C​
PreviousNext

More from Indefinite Integrals

  • If ∫sin3x+cos3xsinx​dx=αloge​∣1+tanx∣+βloge​∣1−tanx+tan2x∣+γtan−1(3​2tanx−1​)+C, when C is constant of…2021 · Numerical
  • If ∫sin−1(1+xx​​)dx = A(x) tan−1(x​) + B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :2020 · MCQ
  • Let f(x)=∫(1+x)2x​​dx(x≥0). Then f(3) – f(1) is eqaul to :2020 · MCQ
  • The integral ∫(xsinx+cosxx​)2dx is equal to (where C is a constant of integration):2020 · MCQ
  • If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c where c is a constant of integration, then g(0) is equal to :2020 · MCQ
  • If ∫5+7sinθ−2cos2θcosθ​dθ= A loge​∣B(θ)∣+C, where C is a constant of integration, then AB(θ)​ can be :2020 · MCQ
  • If ∫sin3x(1+sin6x)2/3cosxdx​=f(x)(1+sin6x)1/λ+c where c is a constant of integration, then λf(3π​)…2020 · MCQ
  • If ƒ'(x) = tan–1(secx + tanx), −2π​<x<2π​, and ƒ(0) = 0, then ƒ(1) is equal to :2020 · MCQ