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Evaluate the following : ∫x3.tan-1x.dx

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Question

Evaluate the following : `int x^3.tan^-1x.dx`

Sum
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Solution

Let I = `int x^3.tan^-1x.dx`

= `int (tan^-1 x).x^3dx`

= `(tan^-1x) int x^3.dx - int [{d/dx (tan^-1 x) int x^3.dx}].dx`

= `(tan^-1x) (x^4/4) - int (1/(1 + x^2))x^4/(4).dx`

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

= `x^4/(4) tan^-1x - (1)/(4) int ((x^2 - 1)(x^2 + 1) + 1)/(x^2 + 1).dx`

= `x^4/(4) tan^-1x - (1)/(4) int [x^2 - 1 + 1/(x^2 + 1)].dx`

= `x^4/(4) tan^-1x - (1)/(4) int [int x^2.dx - int 1.dx + int 1/(x^2 + 1).dx]`

= `x^4/(4) tan^-1x - (1)/(4)[x^3/3 - x + tan^-1x] + c`

= `x^4/(4) tan^-1x - tan^-1  x/(4) - x^3/(12) - x/(4) + c`

= `(1)/(4) (tan^-1x) (x^4 - 1) - x/(12) (x^2 - 3) + c`.

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Chapter 3: Indefinite Integration - Exercise 3.3 [Page 137]

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