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The value of ∫01x4(1-x)41+x2dx is -

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Question

The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is

Options

  • `22/7 - π`

  • `2/105`

  • 0

  • `71/15 - (3π)/2`

MCQ
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Solution

The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is `underlinebb(22/7 - π)`

Explanation:

Let I = `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx`

= `int_0^1 ((x^4 - 1)(1 - x)^4 + (1 - x)^4)/((1 + x^2)) dx`

= `int_0^1 (x^2 - 1)(1 - x)^4 dx + int_0^1 (1 + x^2 - 2x)^2/((1 + x^2)) dx`

= `int_0^1 {(x^2 - 1) (1 - x)^4 + (1 + x^2) - 4x + (4x^2)/((1 + x^2))}dx`

= `int_0^1 {(x^2 - 1) (1 - x)^4 + (1 + x^2) - 4x + 4 - 4/(1 + x^2)}dx`

= `int_0^1(x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - 4/(1 + x^2))dx`

= `[x^7/7 - (4x^6)/6 + (5x^5)/5 - (4x^3)/3 + 4x - 4 tan^-1x]_0^1`

= `1/7 - 4/6 + 5/5 - 4/3 + 4 - 4(π/4 - 0)`

= `22/7 - π`

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