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Indefinite Integrals question

2018 · 15 Apr · Shift 1 · Q44
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  5. /2018 · 15 Apr · Shift 1 · Q44

Indefinite Integrals question

2018 · 15 Apr · Shift 1 · Q44

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If f(x−4x+2)=2x+1,f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,f(x+2x−4​)=2x+1,(x ∈\in∈ R −-−{1, }−-− 2}), then ∫f(x)dx\int f \left( x \right)dx∫f(x)dx is equal to : (where C is a constant of integration)
  1. A
    12 loge | 1 −-− x | + 3x + C
  2. B
    −-− 12 loge | 1 −-− x |−-− 3x + C
  3. C
    12 loge | 1 −-− x |−-− 3x + C
  4. D
    −-− 12 loge | 1 −-− x | + 3x + C
View written solutionFree

Correct answer: B

  1. Given functional equation

We have

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

We want to find f(x)f(x)f(x) first, then integrate it.

  1. Let
t=x−4x+2.t=\frac{x-4}{x+2}.t=x+2x−4​.

Now solve for xxx in terms of ttt.

t(x+2)=x−4t(x+2)=x-4t(x+2)=x−4 tx+2t=x−4tx+2t=x-4tx+2t=x−4 tx−x=−4−2ttx-x=-4-2ttx−x=−4−2t x(t−1)=−(4+2t)x(t-1)=-(4+2t)x(t−1)=−(4+2t) x=−(4+2t)t−1=4+2t1−t.x=\frac{-(4+2t)}{t-1}=\frac{4+2t}{1-t}.x=t−1−(4+2t)​=1−t4+2t​.

So,

x=2(t+2)1−t.x=\frac{2(t+2)}{1-t}.x=1−t2(t+2)​.
  1. Express f(t)f(t)f(t)

Since

f(t)=2x+1,f(t)=2x+1,f(t)=2x+1,

substitute the value of xxx:

f(t)=2(2(t+2)1−t)+1f(t)=2\left(\frac{2(t+2)}{1-t}\right)+1f(t)=2(1−t2(t+2)​)+1 =4(t+2)1−t+1=\frac{4(t+2)}{1-t}+1=1−t4(t+2)​+1 =4t+81−t+1=\frac{4t+8}{1-t}+1=1−t4t+8​+1 =4t+8+1−t1−t=\frac{4t+8+1-t}{1-t}=1−t4t+8+1−t​ =3t+91−t=\frac{3t+9}{1-t}=1−t3t+9​ =3(t+3)1−t.=\frac{3(t+3)}{1-t}.=1−t3(t+3)​.

Thus, replacing ttt by xxx,

f(x)=3x+91−x.f(x)=\frac{3x+9}{1-x}.f(x)=1−x3x+9​.

Now simplify:

3x+91−x=−3x+9x−1.\frac{3x+9}{1-x}=-\frac{3x+9}{x-1}.1−x3x+9​=−x−13x+9​.

Divide:

3x+91−x=−3−12x−1=121−x−3.\frac{3x+9}{1-x}=-3-\frac{12}{x-1}=\frac{12}{1-x}-3.1−x3x+9​=−3−x−112​=1−x12​−3.

Hence,

f(x)=121−x−3.f(x)=\frac{12}{1-x}-3.f(x)=1−x12​−3.
  1. Integrate f(x)f(x)f(x)

We need

∫f(x) dx=∫(121−x−3)dx.\int f(x)\,dx=\int \left(\frac{12}{1-x}-3\right)dx.∫f(x)dx=∫(1−x12​−3)dx.

Now,

∫121−x dx=−12ln⁡∣1−x∣\int \frac{12}{1-x}\,dx = -12\ln|1-x|∫1−x12​dx=−12ln∣1−x∣

because ddx(1−x)=−1\dfrac{d}{dx}(1-x)=-1dxd​(1−x)=−1.

Also,

∫(−3) dx=−3x.\int (-3)\,dx=-3x.∫(−3)dx=−3x.

Therefore,

∫f(x) dx=−12ln⁡∣1−x∣−3x+C.\int f(x)\,dx=-12\ln|1-x|-3x+C.∫f(x)dx=−12ln∣1−x∣−3x+C.
  1. Compare with options

This matches:

B: −12log⁡e∣1−x∣−3x+C\boxed{\text{B: }-12\log_e|1-x|-3x+C}B: −12loge​∣1−x∣−3x+C​
  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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