What is the integral of (x^2)/(sqrt(x^2-81)) ?

The integral of (x^2)/(sqrt(x^2-81)) is 81(1/162 xsqrt(x^2-81)+1/2 ln((|sqrt(x^2-81)+x|)/9))+C

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integral of (x^2)/(sqrt(x^2-81))

\int \frac{x ^{2}}{x ^{2} - 81} d x

81 (\frac{1}{162} x x^{2} - 81 + \frac{1}{2} ln (\frac{x ^{2} - 81 + x}{9})) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 81} d x

Apply Trigonometric Substitution

= \int 81 sec^{3} (u) d u

Take the constant out:

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

= 81 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 81 (\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) ∣

= 81 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{9} x)

= 81 (\frac{sec ^{2} ( arcsec ( \frac{1}{9} x ) ) sin ( arcsec ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{9} x)) + sec (arcsec (\frac{1}{9} x)))

Simplify

81 (\frac{sec ^{2} ( arcsec ( \frac{1}{9} x ) ) sin ( arcsec ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{9} x)) + sec (arcsec (\frac{1}{9} x))) : 81 (\frac{1}{162} x x^{2} - 81 + \frac{1}{2} ln (\frac{x ^{2} - 81 + x}{9}))

= 81 (\frac{1}{162} x x^{2} - 81 + \frac{1}{2} ln (\frac{x ^{2} - 81 + x}{9}))

Add a constant to the solution

= 81 (\frac{1}{162} x x^{2} - 81 + \frac{1}{2} ln (\frac{x ^{2} - 81 + x}{9})) + C

What is the integral of (x^2)/(sqrt(x^2-81)) ?
The integral of (x^2)/(sqrt(x^2-81)) is 81(1/162 xsqrt(x^2-81)+1/2 ln((|sqrt(x^2-81)+x|)/9))+C