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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= 9 \cdot \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 9 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 9 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 9 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{2} + z sin (x yz))

\int \frac{( 1 - u ^{2} - u - u ^{3} )}{1 + u ^{2}} d u

x^{2} - 3

\frac{d}{d z} (ln (\frac{b ^{2} - z ^{2}}{b ^{2} + z ^{2}}))

y^{^{''}} + 9 y = 4 csc^{2} (3 t)

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