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Indefinite Integrals question

2023 · 10 Apr · Shift 2 · Q27
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  5. /2023 · 10 Apr · Shift 2 · Q27

Indefinite Integrals question

2023 · 10 Apr · Shift 2 · Q27

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
For α,β,γ,δ∈N\alpha, \beta, \gamma, \delta \in \mathbb{N}α,β,γ,δ∈N, if ∫((xe)2x+(ex)2x)log⁡exdx=1α(xe)βx−1γ(ex)δx+C\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C∫((ex​)2x+(xe​)2x)loge​xdx=α1​(ex​)βx−γ1​(xe​)δx+C, where e=∑n=0∞1n!e=\sum_{n=0}^{\infty} \frac{1}{n !}e=∑n=0∞​n!1​ and C\mathrm{C}C is constant of integration, then α+2β+3γ−4δ\alpha+2 \beta+3 \gamma-4 \deltaα+2β+3γ−4δ is equal to :
  1. A
    −8-8−8
  2. B
    −4-4−4
  3. C
    1
  4. D
    4
View written solutionFree

Correct answer: D

  1. Given integral

We need to evaluate

∫[(xe)2x+(ex)2x]ln⁡x dx\int \left[\left(\frac{x}{e}\right)^{2x}+\left(\frac{e}{x}\right)^{2x}\right]\ln x\,dx∫[(ex​)2x+(xe​)2x]lnxdx

and compare it with

1α(xe)βx−1γ(ex)δx+C.\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C.α1​(ex​)βx−γ1​(xe​)δx+C.

We must identify α,β,γ,δ∈N\alpha,\beta,\gamma,\delta\in\mathbb Nα,β,γ,δ∈N.

  1. Differentiate likely forms

Let

y1=(xe)2x.y_1=\left(\frac{x}{e}\right)^{2x}.y1​=(ex​)2x.

Using logarithmic differentiation,

ln⁡y1=2xln⁡(xe)=2x(ln⁡x−1).\ln y_1=2x\ln\left(\frac{x}{e}\right)=2x(\ln x-1).lny1​=2xln(ex​)=2x(lnx−1).

Differentiate:

y1′y1=2(ln⁡x−1)+2x⋅1x=2ln⁡x.\frac{y_1'}{y_1}=2(\ln x-1)+2x\cdot \frac{1}{x}=2\ln x.y1​y1′​​=2(lnx−1)+2x⋅x1​=2lnx.

Hence,

ddx(xe)2x=2ln⁡x(xe)2x.\frac{d}{dx}\left(\frac{x}{e}\right)^{2x}=2\ln x\left(\frac{x}{e}\right)^{2x}.dxd​(ex​)2x=2lnx(ex​)2x.

Therefore,

\left(\frac{x}{e}\right)^{2x}\ln x= rac12\frac{d}{dx}\left(\frac{x}{e}\right)^{2x}.
  1. Now for the second term

Let

y2=(ex)2x.y_2=\left(\frac{e}{x}\right)^{2x}.y2​=(xe​)2x.

Then

ln⁡y2=2xln⁡(ex)=2x(1−ln⁡x).\ln y_2=2x\ln\left(\frac{e}{x}\right)=2x(1-\ln x).lny2​=2xln(xe​)=2x(1−lnx).

Differentiate:

y2′y2=2(1−ln⁡x)+2x(−1x)=−2ln⁡x.\frac{y_2'}{y_2}=2(1-\ln x)+2x\left(-\frac{1}{x}\right)=-2\ln x.y2​y2′​​=2(1−lnx)+2x(−x1​)=−2lnx.

So,

ddx(ex)2x=−2ln⁡x(ex)2x.\frac{d}{dx}\left(\frac{e}{x}\right)^{2x}=-2\ln x\left(\frac{e}{x}\right)^{2x}.dxd​(xe​)2x=−2lnx(xe​)2x.

Hence,

(ex)2xln⁡x=−12ddx(ex)2x.\left(\frac{e}{x}\right)^{2x}\ln x=-\frac12\frac{d}{dx}\left(\frac{e}{x}\right)^{2x}.(xe​)2xlnx=−21​dxd​(xe​)2x.
  1. Integrate termwise

So the integrand becomes

(xe)2xln⁡x+(ex)2xln⁡x=12ddx(xe)2x−12ddx(ex)2x.\left(\frac{x}{e}\right)^{2x}\ln x+\left(\frac{e}{x}\right)^{2x}\ln x =\frac12\frac{d}{dx}\left(\frac{x}{e}\right)^{2x}-\frac12\frac{d}{dx}\left(\frac{e}{x}\right)^{2x}.(ex​)2xlnx+(xe​)2xlnx=21​dxd​(ex​)2x−21​dxd​(xe​)2x.

Integrating,

∫[(xe)2x+(ex)2x]ln⁡x dx=12(xe)2x−12(ex)2x+C.\int \left[\left(\frac{x}{e}\right)^{2x}+\left(\frac{e}{x}\right)^{2x}\right]\ln x\,dx =\frac12\left(\frac{x}{e}\right)^{2x}-\frac12\left(\frac{e}{x}\right)^{2x}+C.∫[(ex​)2x+(xe​)2x]lnxdx=21​(ex​)2x−21​(xe​)2x+C.

Comparing with

1α(xe)βx−1γ(ex)δx+C,\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C,α1​(ex​)βx−γ1​(xe​)δx+C,

we get

α=2,β=2,γ=2,δ=2.\alpha=2,\quad \beta=2,\quad \gamma=2,\quad \delta=2.α=2,β=2,γ=2,δ=2.
  1. Compute the required expression
α+2β+3γ−4δ=2+2(2)+3(2)−4(2).\alpha+2\beta+3\gamma-4\delta =2+2(2)+3(2)-4(2).α+2β+3γ−4δ=2+2(2)+3(2)−4(2). =2+4+6−8=4.=2+4+6-8=4.=2+4+6−8=4.
  1. Option check

The value is

4,4,4,

which corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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