What is the integral of (14)/((1-x^2)^{3/2)} ?

The integral of (14)/((1-x^2)^{3/2)} is (14x)/((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 (14)/((1-x^2)^{3/2)}

\int \frac{14}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{14 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{14}{( 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

= 14 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 14 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 14 tan (u)

Substitute back

u = arcsin (x)

= 14 tan (arcsin (x))

Simplify

14 tan (arcsin (x)) : \frac{14 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{14 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{14 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

\int ln (x^{6}) d x

\frac{d y}{d t} = \frac{3 ( y + 1 )}{t ^{2} + 5 t + 6}

\frac{d y}{d x} = \frac{x ^{4} y ^{4} + 4 x ^{4}}{y ^{3}}

\int \frac{1}{x ^{3} 4 x ^{2} - 9} d x

What is the integral of (14)/((1-x^2)^{3/2)} ?
The integral of (14)/((1-x^2)^{3/2)} is (14x)/((1-x^2)^{1/2)}+C