What is the integral of 1/((x^2-441)^{3/2)} ?

The integral of 1/((x^2-441)^{3/2)} is -x/(441(x^2-441)^{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 1/((x^2-441)^{3/2)}

\int \frac{1}{( x ^{2} - 441 ) ^{\frac{3}{2}}} d x

- \frac{x}{441 ( x ^{2} - 441 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 441 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{441 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{441 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{441} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{441} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{441} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{441} (- \frac{1}{v})

Substitute back

= \frac{1}{441} (- \frac{1}{sin ( arcsec ( \frac{1}{21} x ) )})

Simplify

\frac{1}{441} (- \frac{1}{sin ( arcsec ( \frac{1}{21} x ) )}) : - \frac{x}{441 ( x ^{2} - 441 ) ^{\frac{1}{2}}}

= - \frac{x}{441 ( x ^{2} - 441 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{441 ( x ^{2} - 441 ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (3 (3 x^{2} + 5 x - 1)^{2})

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{- 2 t}

\frac{2 s - 6}{s ^{2} + 9}

\frac{d}{d x} (sin^{4} (2 x co) s^{4} 3 x)

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

What is the integral of 1/((x^2-441)^{3/2)} ?
The integral of 1/((x^2-441)^{3/2)} is -x/(441(x^2-441)^{1/2)}+C