Question

The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.

विकल्प

• – x + (1 + x²) tan^–1x + c

• x – (1 + x²) cot^–1x + c

• – x + (1 + x²) cot^–1x + c

• x – (1 + x²) tan^–1x + c

Answer 1

The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to `underlinebb(-x + (1 + x^2) tan^-1x + c)`.

Explanation:

Let I = `int x cos^-1 ((1 - x^2)/(1 + x^2))dx`

∴ I = \[\ce{2 \int\underset{II}{x}. \underset{I}{tan}^{-1}x dx}\]

Applying Integration by parts

I = `2[tan^-1x int xdx - int(d/(dx) (tan^-1 x) intxdx)dx]`

I = `2[x^2/2 tan^-1 x - int 1/(1 + x^2) xx x^2/2 dx] + c`

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

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

I = `2[x^2/2 tan^-1 x - 1/2 int 1.dx + 1/2 tan^-1 x] + c`

I = `2[x^2/2 tan^-1 x - x/2 + 1/2 tan^-1 x] + c`

I = x² tan^–1 x + tan^–1 x – x + c

or I = –x + (x² + 1) tan^–1 x + c