What is the integral of 5/(x^2sqrt(x^2-9)) ?

The integral of 5/(x^2sqrt(x^2-9)) is (5sqrt(x^2-9))/(9x)+C

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integral of 5/(x^2sqrt(x^2-9))

\int \frac{5}{x ^{2} x ^{2} - 9} d x

\frac{5 x ^{2} - 9}{9 x} + C

Solution steps

\int \frac{5}{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

= 5 \cdot \int \frac{1}{x ^{2} x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( u )} d u

= 5 \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

= 5 \cdot \frac{1}{9} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= 5 \cdot \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= 5 \cdot \frac{1}{9} sin (u)

Substitute back

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

= 5 \cdot \frac{1}{9} sin (arcsec (\frac{1}{3} x))

Simplify

5 \cdot \frac{1}{9} sin (arcsec (\frac{1}{3} x)) : \frac{5 x ^{2} - 9}{9 x}

= \frac{5 x ^{2} - 9}{9 x}

Add a constant to the solution

= \frac{5 x ^{2} - 9}{9 x} + C

\frac{d}{d x} (x e^{- x^{2} y})

\frac{d ^{2}}{d x ^{2}} (3 x sin (x^{2}))

\int \frac{x}{x - 9} d x

\int \frac{1}{x x ^{3} - 1} d x

\int \frac{- 3}{x ^{2}} + 11 - 4^{x} d x

What is the integral of 5/(x^2sqrt(x^2-9)) ?
The integral of 5/(x^2sqrt(x^2-9)) is (5sqrt(x^2-9))/(9x)+C