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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/((1-x^2)^{3/2)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec^{2} (u) d u

= 2 \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 2 tan (u)

Substitute back

u = arcsin (x)

= 2 tan (arcsin (x))

Simplify

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

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

Add a constant to the solution

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

\frac{1}{( s \cdot ( s + 1 ) )}

\int \frac{x ^{2}}{16 + 9 x ^{2}} d x

\int_{1}^{2} 2 x - x^{2} d x

\int \frac{1}{x ^{2} - 2 x + 37} d x

\frac{d x}{d t} + x^{2} = x

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