What is the integral of (72)/(x^2sqrt(4x^2+9)) ?

The integral of (72)/(x^2sqrt(4x^2+9)) is -(8sqrt(4x^2+9))/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (72)/(x^2sqrt(4x^2+9))

\int \frac{72}{x ^{2} 4 x ^{2} + 9} d x

- \frac{8 4 x ^{2} + 9}{x} + C

Solution steps

\int \frac{72}{x ^{2} 4 x ^{2} + 9} d x

Take the constant out:

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

= 72 \cdot \int \frac{1}{x ^{2} 4 x ^{2} + 9} d x

Apply Trigonometric Substitution

: \int \frac{2 sec ( u )}{9 tan ^{2} ( u )} d u

= 72 \cdot \int \frac{2 sec ( u )}{9 tan ^{2} ( u )} d u

Take the constant out:

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

= 72 \cdot \frac{2}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 72 \cdot \frac{2}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 72 \cdot \frac{2}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 72 \cdot \frac{2}{9} (- \frac{1}{v})

Substitute back

= 72 \cdot \frac{2}{9} (- \frac{1}{sin ( arctan ( \frac{2}{3} x ) )})

Simplify

72 \cdot \frac{2}{9} (- \frac{1}{sin ( arctan ( \frac{2}{3} x ) )}) : - \frac{8 4 x ^{2} + 9}{x}

= - \frac{8 4 x ^{2} + 9}{x}

Add a constant to the solution

= - \frac{8 4 x ^{2} + 9}{x} + C

n = 0 \sum \infty (5 x)^{n}

\frac{1}{1 - x ^{2}}

y = x, (4, 2)

\frac{d}{d x} (3 x^{2} + 12 x + 9)

\int \frac{1}{36 e ^{- 5 x} + e ^{5 x}} d x

What is the integral of (72)/(x^2sqrt(4x^2+9)) ?
The integral of (72)/(x^2sqrt(4x^2+9)) is -(8sqrt(4x^2+9))/x+C