What is the integral of 1/(2sqrt(3-x)\sqrt{x+1)} ?

The integral of 1/(2sqrt(3-x)\sqrt{x+1)} is arcsin(1/2 sqrt(x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2sqrt(3-x)\sqrt{x+1)}

\int \frac{1}{2 3 - x x + 1} d x

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

Solution steps

\int \frac{1}{2 3 - x x + 1} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{3 - x x + 1} d x

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

3 - x x + 1 = (3 - x) (x + 1)

= \frac{1}{2} \cdot \int \frac{1}{( 3 - x ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{1}{u ( - u + 4 )} d u

= \frac{1}{2} \cdot \int \frac{1}{u ( - u + 4 )} d u

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

u (- u + 4) = u (- u + 4)

= \frac{1}{2} \cdot \int \frac{1}{u - u + 4} d u

Apply u-substitution

: \int \frac{2}{- v ^{2} + 4} d v

= \frac{1}{2} \cdot \int \frac{2}{- v ^{2} + 4} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{- v ^{2} + 4} d v

Apply Trigonometric Substitution

: \int 1 d w

= \frac{1}{2} \cdot 2 \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 2 \cdot 1 \cdot w

Substitute back

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

Simplify

\frac{1}{2} \cdot 2 \cdot 1 \cdot arcsin (\frac{1}{2} x + 1) : arcsin (\frac{1}{2} x + 1)

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x^{2} e^{x}))

n = 0 \sum \infty \frac{1}{( n ^{3} - n )}

y^{^{'}} + 8 x - 1 y = x^{8}, y (1) = 9

p p

\int \frac{( 3 x + 1 ) ^{2}}{244} d x

What is the integral of 1/(2sqrt(3-x)\sqrt{x+1)} ?
The integral of 1/(2sqrt(3-x)\sqrt{x+1)} is arcsin(1/2 sqrt(x+1))+C