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Indefinite Integrals question

2019 · 12 Apr · Shift 1 · Q28
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  5. /2019 · 12 Apr · Shift 1 · Q28

Indefinite Integrals question

2019 · 12 Apr · Shift 1 · Q28

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
The integral ∫2x3−1x4+xdx\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx∫x4+x2x3−1​dx is equal to : (Here C is a constant of integration)
  1. A
    log⁡e∣x3+1∣x2+C{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + Cloge​x2​x3+1​​+C
  2. B
    12log⁡e∣x3+1∣x2+C{1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C21​loge​x2​x3+1​​+C
  3. C
    log⁡e∣x3+1x∣+C{\log _e}\left| {{{{x^3} + 1} \over x}} \right| + Cloge​​xx3+1​​+C
  4. D
    12log⁡e(x3+1)2∣x3∣+C{1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C21​loge​∣x3∣(x3+1)2​+C
View written solutionFree

Correct answer: C

  1. Simplify the integrand

We need to evaluate

I=∫2x3−1x4+x dx.I=\int \frac{2x^3-1}{x^4+x}\,dx.I=∫x4+x2x3−1​dx.

Factor the denominator:

x4+x=x(x3+1).x^4+x=x(x^3+1).x4+x=x(x3+1).

So,

I=∫2x3−1x(x3+1) dx.I=\int \frac{2x^3-1}{x(x^3+1)}\,dx.I=∫x(x3+1)2x3−1​dx.

Now observe that

2x3−1=2(x3+1)−3.2x^3-1=2(x^3+1)-3.2x3−1=2(x3+1)−3.

Hence,

2x3−1x(x3+1)=2(x3+1)−3x(x3+1)=2x−3x(x3+1).\frac{2x^3-1}{x(x^3+1)} =\frac{2(x^3+1)-3}{x(x^3+1)} =\frac{2}{x}-\frac{3}{x(x^3+1)}.x(x3+1)2x3−1​=x(x3+1)2(x3+1)−3​=x2​−x(x3+1)3​.

A cleaner way is to compare with the derivative of

ln⁡∣x3+1x∣.\ln\left|\frac{x^3+1}{x}\right|.ln​xx3+1​​.
  1. Differentiate the likely answer form

Let

f(x)=ln⁡∣x3+1x∣=ln⁡∣x3+1∣−ln⁡∣x∣.f(x)=\ln\left|\frac{x^3+1}{x}\right|=\ln|x^3+1|-\ln|x|.f(x)=ln​xx3+1​​=ln∣x3+1∣−ln∣x∣.

Then,

f′(x)=3x2x3+1−1x.f'(x)=\frac{3x^2}{x^3+1}-\frac{1}{x}.f′(x)=x3+13x2​−x1​.

Taking LCM:

f′(x)=3x3−(x3+1)x(x3+1)=2x3−1x(x3+1).f'(x)=\frac{3x^3-(x^3+1)}{x(x^3+1)} =\frac{2x^3-1}{x(x^3+1)}.f′(x)=x(x3+1)3x3−(x3+1)​=x(x3+1)2x3−1​.

Since

x(x3+1)=x4+x,x(x^3+1)=x^4+x,x(x3+1)=x4+x,

we get

f′(x)=2x3−1x4+x.f'(x)=\frac{2x^3-1}{x^4+x}.f′(x)=x4+x2x3−1​.

Therefore,

∫2x3−1x4+x dx=ln⁡∣x3+1x∣+C.\int \frac{2x^3-1}{x^4+x}\,dx = \ln\left|\frac{x^3+1}{x}\right|+C.∫x4+x2x3−1​dx=ln​xx3+1​​+C.
  1. Match with the options

This is exactly Option C:

log⁡e∣x3+1x∣+C.\log_e\left|\frac{x^3+1}{x}\right|+C.loge​​xx3+1​​+C.
  1. Check other equivalent-looking options

Option A:

log⁡e(∣x3+1∣x2)+C=ln⁡∣x3+1∣−2ln⁡∣x∣+C,\log_e\left(\frac{|x^3+1|}{x^2}\right)+C =\ln|x^3+1|-2\ln|x|+C,loge​(x2∣x3+1∣​)+C=ln∣x3+1∣−2ln∣x∣+C,

whose derivative is

3x2x3+1−2x,\frac{3x^2}{x^3+1}-\frac{2}{x},x3+13x2​−x2​,

not equal to the integrand.

Option B is half of A, so also incorrect.

Option D:

12ln⁡((x3+1)2∣x3∣)+C=ln⁡∣x3+1∣−12ln⁡∣x3∣+C=ln⁡∣x3+1∣−32ln⁡∣x∣+C,\frac12\ln\left(\frac{(x^3+1)^2}{|x^3|}\right)+C =\ln|x^3+1|-\frac12\ln|x^3|+C =\ln|x^3+1|-\frac32\ln|x|+C,21​ln(∣x3∣(x3+1)2​)+C=ln∣x3+1∣−21​ln∣x3∣+C=ln∣x3+1∣−23​ln∣x∣+C,

whose derivative is

3x2x3+1−32x,\frac{3x^2}{x^3+1}-\frac{3}{2x},x3+13x2​−2x3​,

again not equal to the integrand.

Hence only C is correct.

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