What is the integral of 1/(x^2sqrt(x^2+9)) ?

The integral of 1/(x^2sqrt(x^2+9)) is -(sqrt(9+x^2))/(9x)+C

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integral of 1/(x^2sqrt(x^2+9))

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to 3 lim (7 x + ∣ x - 3 ∣)

\frac{\partial}{\partial x} (e^{y} + x e^{x y})

x^{3} f (x)

\int_{3}^{6} 4 d x

\frac{\partial}{\partial x} ((x + 2)^{2} - 2 (y - 1)^{2} - 5)

What is the integral of 1/(x^2sqrt(x^2+9)) ?
The integral of 1/(x^2sqrt(x^2+9)) is -(sqrt(9+x^2))/(9x)+C