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{1}{9 sec ( u )} d u

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

Rewrite using trig identities

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

Use the common integral:

\int cos (u) d u = sin (u)

= 6 \cdot \frac{1}{9} sin (u)

Substitute back

u = arcsec (\frac{1}{3} x)

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

Simplify

6 \cdot \frac{1}{9} sin (arcsec (\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

\frac{d}{d x} (5^{x - 2})

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 4

\frac{d}{d x} (\frac{( 2 x ^{2} + 8 x + 6 )}{x})

cos (\frac{1 - e ^{3 x}}{1 + e ^{3 x}})

\frac{\partial}{\partial x} (2 x y^{3} - 4 x^{2} y^{2})

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