What is the integral of 6/(x^2sqrt(x^2+9)) ?

The integral of 6/(x^2sqrt(x^2+9)) is -(2sqrt(x^2+9))/(3x)+C

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integral of 6/(x^2sqrt(x^2+9))

\int \frac{6}{x ^{2} x ^{2} + 9} d x

- \frac{2 x ^{2} + 9}{3 x} + C

Solution steps

\int \frac{6}{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

= 6 \cdot \int \frac{1}{x ^{2} x ^{2} + 9} d x

Apply Trigonometric Substitution

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

= 6 \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

= 6 \cdot \frac{1}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 6 \cdot \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 6 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 6 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 6 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )})

Simplify

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

= - \frac{2 x ^{2} + 9}{3 x}

Add a constant to the solution

= - \frac{2 x ^{2} + 9}{3 x} + C

d x = e^{3 x} d y

\int s 8^{s} d s

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

y = 3 - 6 x

\frac{\partial}{\partial x} (x^{2} cos (x))

What is the integral of 6/(x^2sqrt(x^2+9)) ?
The integral of 6/(x^2sqrt(x^2+9)) is -(2sqrt(x^2+9))/(3x)+C