Consider `int x tan^(-1) x dx = A (x^(2) + 1) tan^(-1) x + Bx + C`, where C is the constant of integration
What is the value of B ?

Options

A. 1
B. `1//2`
C. `-1//2`
D. `1//4`

Answer

C

Explanation

Given, `int x tan ^(-1) x dx = A(x^(2) + 1) tan^(-1) x + Bx + C`
where, C is the constant of integration
Consider, `int underset(II)(x) underset(I)(tan)^(-1) x dx`
`= tan^(-1) x.(x^(2))/(2) - int (d)/(dx) (tan^(-1)x) .(x^(2))/(2) dx` (using integration by parts)
`= (x^(2).tan^(-1)x)/(2) - (1)/(2) int (x^(2))/(1 + x^(2)) dx`
`= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int ((1 + x^(2) -1)/(1 + x^(2)))dx)`
`= (x^(2) tan^(-1) x)/(2) - (1)/(2) (int dx - int (dx)/(1 + x^(2)))`
`= (x^(2) tan^(-1) x)/(2) - (1)/(2) (x - tan^(-1) x) + C`
`= (x^(2) tan^(-1)x)/(2) - (x)/(2) + (tan^(-1)x)/(2) + C`
`= (1)/(2) (x^(2) + 1 ) tan^(-1) x - (x)/(2) + C`
`B = - (1)/(2)`, hence option (c) is correct