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

The integral of (x^2)/(sqrt(x^2-9)) is 9(1/18 xsqrt(x^2-9)+1/2 ln((|sqrt(x^2-9)+x|)/3))+C

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

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

9 (\frac{1}{18} x x^{2} - 9 + \frac{1}{2} ln (\frac{x ^{2} - 9 + x}{3})) + C

Solution steps

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

Apply Trigonometric Substitution

= \int 9 sec^{3} (u) d u

Take the constant out:

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

= 9 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= 9 (\frac{sec ^{2} ( arcsec ( \frac{1}{3} x ) ) sin ( arcsec ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)))

Simplify

9 (\frac{sec ^{2} ( arcsec ( \frac{1}{3} x ) ) sin ( arcsec ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))) : 9 (\frac{1}{18} x x^{2} - 9 + \frac{1}{2} ln (\frac{x ^{2} - 9 + x}{3}))

= 9 (\frac{1}{18} x x^{2} - 9 + \frac{1}{2} ln (\frac{x ^{2} - 9 + x}{3}))

Add a constant to the solution

= 9 (\frac{1}{18} x x^{2} - 9 + \frac{1}{2} ln (\frac{x ^{2} - 9 + x}{3})) + C

What is the integral of (x^2)/(sqrt(x^2-9)) ?
The integral of (x^2)/(sqrt(x^2-9)) is 9(1/18 xsqrt(x^2-9)+1/2 ln((|sqrt(x^2-9)+x|)/3))+C