What is the integral of (10)/(x^2sqrt(x^2+9)) ?

The integral of (10)/(x^2sqrt(x^2+9)) is -(10sqrt(9+x^2))/(9x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (10)/(x^2sqrt(x^2+9))

\int \frac{10}{x ^{2} x ^{2} + 9} d x

- \frac{10 9 + x ^{2}}{9 x} + C

Solution steps

\int \frac{10}{x ^{2} x ^{2} + 9} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{x ^{2} x ^{2} + 9} d x

Apply Trigonometric Substitution

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

= 10 \cdot \int \frac{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

= 10 \cdot \frac{1}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 10 \cdot \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 10 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 10 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 10 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )})

Simplify

10 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )}) : - \frac{10 9 + x ^{2}}{9 x}

= - \frac{10 9 + x ^{2}}{9 x}

Add a constant to the solution

= - \frac{10 9 + x ^{2}}{9 x} + C

y = 565 + t (37 t^{0.6} - 103)

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x e^{- 2 x}

\frac{d q}{d t} = k q^{2}

f (x) = (3 x + 1)^{5}

What is the integral of (10)/(x^2sqrt(x^2+9)) ?
The integral of (10)/(x^2sqrt(x^2+9)) is -(10sqrt(9+x^2))/(9x)+C