What is the integral of (sqrt(1-x^2))/(1+x^2) ?

The integral of (sqrt(1-x^2))/(1+x^2) is sqrt(2)arctan((sqrt(2)x)/(sqrt(1-x^2)))-arcsin(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(1-x^2))/(1+x^2)

\int \frac{1 - x ^{2}}{1 + x ^{2}} d x

2 arctan (\frac{2 x}{1 - x ^{2}}) - arcsin (x) + C

Solution steps

\int \frac{1 - x ^{2}}{1 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cos ^{2} ( u )}{1 + sin ^{2} ( u )} d u

= \int \frac{cos ^{2} ( u )}{1 + sin ^{2} ( u )} d u

Rewrite using trig identities

= \int \frac{1 - sin ^{2} ( u )}{1 + sin ^{2} ( u )} d u

\frac{1 - sin ^{2} ( u )}{1 + sin ^{2} ( u )} = \frac{2}{1 + sin ^{2} ( u )} - 1

= \int \frac{2}{1 + sin ^{2} ( u )} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2}{1 + sin ^{2} ( u )} d u - \int 1 d u

\int \frac{2}{1 + sin ^{2} ( u )} d u = 2 arctan (2 tan (u))

\int 1 d u = u

= 2 arctan (2 tan (u)) - u

Substitute back

u = arcsin (x)

= 2 arctan (2 tan (arcsin (x))) - arcsin (x)

Simplify

2 arctan (2 tan (arcsin (x))) - arcsin (x) : 2 arctan (\frac{2 x}{1 - x ^{2}}) - arcsin (x)

= 2 arctan (\frac{2 x}{1 - x ^{2}}) - arcsin (x)

Add a constant to the solution

= 2 arctan (\frac{2 x}{1 - x ^{2}}) - arcsin (x) + C

\int \frac{x}{25 + 3 x} d x

\int e^{- l x} d x

\int \frac{1}{- 4 x ^{2} - 12 x - 8} d x

\int \frac{12 x ^{2} - 4 x}{( x ^{2} + 4 ) ( x + 2 )} d x

What is the integral of (sqrt(1-x^2))/(1+x^2) ?
The integral of (sqrt(1-x^2))/(1+x^2) is sqrt(2)arctan((sqrt(2)x)/(sqrt(1-x^2)))-arcsin(x)+C