Question

${\pink{ \boxed{ ∫ {x}^{3}. \ { \tan }^{ - 1} x \: dx}}} \}$


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Answer 1

Answer:

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

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

- 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.

Answer 2

Refer to attachment

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