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Indefinite Integrals question

2021 · 25 Feb · Shift 2 · Q27
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  5. /2021 · 25 Feb · Shift 2 · Q27

Indefinite Integrals question

2021 · 25 Feb · Shift 2 · Q27

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
The integral ∫e3log⁡e2x+5e2log⁡e2xe4log⁡ex+5e3log⁡ex−7e2log⁡exdx\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx∫e4loge​x+5e3loge​x−7e2loge​xe3loge​2x+5e2loge​2x​dx, x > 0, is equal to : (where c is a constant of integration)
  1. A
    log⁡ex2+5x−7+c{\log _e}\sqrt {{x^2} + 5x - 7} + cloge​x2+5x−7​+c
  2. B
    4log⁡e∣x2+5x−7∣+c4{\log _e}|{x^2} + 5x - 7| + c4loge​∣x2+5x−7∣+c
  3. C
    14log⁡e∣x2+5x−7∣+c{1 \over 4}{\log _e}|{x^2} + 5x - 7| + c41​loge​∣x2+5x−7∣+c
  4. D
    log⁡e∣x2+5x−7∣+c{\log _e}|{x^2} + 5x - 7| + cloge​∣x2+5x−7∣+c
View written solutionFree

Correct answer: B

  1. Rewrite the exponential terms using elog⁡ea=ae^{\log_e a}=aeloge​a=a.

Given

I=∫e3log⁡e(2x)+5e2log⁡e(2x)e4log⁡ex+5e3log⁡ex−7e2log⁡ex dx,x>0I=\int \frac{e^{3\log_e(2x)}+5e^{2\log_e(2x)}}{e^{4\log_e x}+5e^{3\log_e x}-7e^{2\log_e x}}\,dx, \qquad x>0I=∫e4loge​x+5e3loge​x−7e2loge​xe3loge​(2x)+5e2loge​(2x)​dx,x>0

Since x>0x>0x>0,

e3log⁡e(2x)=(2x)3=8x3,e^{3\log_e(2x)}=(2x)^3=8x^3,e3loge​(2x)=(2x)3=8x3, e2log⁡e(2x)=(2x)2=4x2,e^{2\log_e(2x)}=(2x)^2=4x^2,e2loge​(2x)=(2x)2=4x2, e4log⁡ex=x4,e3log⁡ex=x3,e2log⁡ex=x2.e^{4\log_e x}=x^4,\quad e^{3\log_e x}=x^3,\quad e^{2\log_e x}=x^2.e4loge​x=x4,e3loge​x=x3,e2loge​x=x2.

So the integrand becomes

8x3+5⋅4x2x4+5x3−7x2=8x3+20x2x4+5x3−7x2.\frac{8x^3+5\cdot 4x^2}{x^4+5x^3-7x^2} =\frac{8x^3+20x^2}{x^4+5x^3-7x^2}.x4+5x3−7x28x3+5⋅4x2​=x4+5x3−7x28x3+20x2​.
  1. Factor numerator and denominator.
8x3+20x2=4x2(2x+5),8x^3+20x^2=4x^2(2x+5),8x3+20x2=4x2(2x+5), x4+5x3−7x2=x2(x2+5x−7).x^4+5x^3-7x^2=x^2(x^2+5x-7).x4+5x3−7x2=x2(x2+5x−7).

Hence

I=∫4x2(2x+5)x2(x2+5x−7) dx=∫4(2x+5)x2+5x−7 dx.I=\int \frac{4x^2(2x+5)}{x^2(x^2+5x-7)}\,dx =\int \frac{4(2x+5)}{x^2+5x-7}\,dx.I=∫x2(x2+5x−7)4x2(2x+5)​dx=∫x2+5x−74(2x+5)​dx.
  1. Recognize derivative of denominator.

Let

u=x2+5x−7.u=x^2+5x-7.u=x2+5x−7.

Then

dνdx=2x+5.\frac{d\nu}{dx}=2x+5.dxdν​=2x+5.

Therefore,

I=∫42x+5x2+5x−7 dx=4∫dνν.I=\int 4\frac{2x+5}{x^2+5x-7}\,dx =4\int \frac{d\nu}{\nu}.I=∫4x2+5x−72x+5​dx=4∫νdν​.
  1. Integrate.
I=4ln⁡∣ν∣+c=4ln⁡∣x2+5x−7∣+c.I=4\ln|\nu|+c=4\ln|x^2+5x-7|+c.I=4ln∣ν∣+c=4ln∣x2+5x−7∣+c.
  1. Match with the options.

This is exactly Option B:

4log⁡e∣x2+5x−7∣+c.4\log_e|x^2+5x-7|+c.4loge​∣x2+5x−7∣+c.
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