What is the integral of ((x+2))/(sqrt(4-x^2)) ?

The integral of ((x+2))/(sqrt(4-x^2)) is -sqrt(4-x^2)+2arcsin(1/2 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+2))/(sqrt(4-x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 2 (sin (u) + 1) d u

= \int 2 (sin (u) + 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin (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

= 2 (\int sin (u) d u + \int 1 d u)

\int sin (u) d u = - cos (u)

\int 1 d u = u

= 2 (- cos (u) + u)

Substitute back

u = arcsin (\frac{1}{2} x)

= 2 (- cos (arcsin (\frac{1}{2} x)) + arcsin (\frac{1}{2} x))

Simplify

2 (- cos (arcsin (\frac{1}{2} x)) + arcsin (\frac{1}{2} x)) : - 4 - x^{2} + 2 arcsin (\frac{1}{2} x)

= - 4 - x^{2} + 2 arcsin (\frac{1}{2} x)

Add a constant to the solution

= - 4 - x^{2} + 2 arcsin (\frac{1}{2} x) + C

\int sec^{2} (\frac{x}{6}) d x

\int sin^{4} (\frac{x}{9}) cos (\frac{x}{9}) d x

\frac{1}{( x ^{2} + 1 )}

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

\int csc^{2} (x + 1) d x

What is the integral of ((x+2))/(sqrt(4-x^2)) ?
The integral of ((x+2))/(sqrt(4-x^2)) is -sqrt(4-x^2)+2arcsin(1/2 x)+C