If ∫2e5x+e4x-4e3x+4e2x+2ex(e2x+4)(e2x-1)2dx=tan-1(exa)-1b(e2x-1)+C, where C is constant of integration, then value of a + b is equal to ______. - Mathematics (JEE Main)

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MCQ
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If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.

Options

  • 2.00

  • 3.00

  • 4.00

  • 5.00

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Solution

If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to 4.00.

Explanation:

I = `int{(2(e^x)^4 + (e^x)^3 - 4(e^x)^2 + 4e^x + 2)/(((e^x)^2 + 4)((e^x)^2 - 1)^2)}d(e^x)`

Put ex = t

I = `int ((2t^4 - 4t^2 + 2) + (t^3 + 4t))/((t^2 + 4)(t^2 - 1)^2)dt`

= `int 2/(t^2 + 4) + t/(t^2 - 1)^2 dt`

= `tan^-1(t/2) - 1/(2(t^2 - 1)) + c`

∴ a + b = 4

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