What is the integral of 7/(x^2sqrt(x^2+9)) ?

The integral of 7/(x^2sqrt(x^2+9)) is -(7sqrt(9+x^2))/(9x)+C

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integral of 7/(x^2sqrt(x^2+9))

\int \frac{7}{x ^{2} x ^{2} + 9} d x

- \frac{7 9 + x ^{2}}{9 x} + C

Solution steps

\int \frac{7}{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

= 7 \cdot \int \frac{1}{x ^{2} x ^{2} + 9} d x

Apply Trigonometric Substitution

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

= 7 \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

= 7 \cdot \frac{1}{9} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 7 \cdot \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 7 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 7 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 7 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )})

Simplify

7 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} x ) )}) : - \frac{7 9 + x ^{2}}{9 x}

= - \frac{7 9 + x ^{2}}{9 x}

Add a constant to the solution

= - \frac{7 9 + x ^{2}}{9 x} + C

\int \frac{1}{y ( 1 - y )} d y

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

(2 + x) y^{^{'}} = 8 y

x \to 0 lim (e^{x} - e^{- x})

\frac{d y}{d x} = 9 e^{x - y}

What is the integral of 7/(x^2sqrt(x^2+9)) ?
The integral of 7/(x^2sqrt(x^2+9)) is -(7sqrt(9+x^2))/(9x)+C