What is the integral of (x^2)/(sqrt(x^2-16)) ?

The integral of (x^2)/(sqrt(x^2-16)) is 16(1/32 xsqrt(x^2-16)+1/2 ln((|sqrt(x^2-16)+x|)/4))+C

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integral of (x^2)/(sqrt(x^2-16))

\int \frac{x ^{2}}{x ^{2} - 16} d x

16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4})) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 16} d x

Apply Trigonometric Substitution

= \int 16 sec^{3} (u) d u

Take the constant out:

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

= 16 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 16 (\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) ∣

= 16 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{4} x)

= 16 (\frac{sec ^{2} ( arcsec ( \frac{1}{4} x ) ) sin ( arcsec ( \frac{1}{4} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x)))

Simplify

16 (\frac{sec ^{2} ( arcsec ( \frac{1}{4} x ) ) sin ( arcsec ( \frac{1}{4} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x))) : 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4}))

= 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4}))

Add a constant to the solution

= 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4})) + C

What is the integral of (x^2)/(sqrt(x^2-16)) ?
The integral of (x^2)/(sqrt(x^2-16)) is 16(1/32 xsqrt(x^2-16)+1/2 ln((|sqrt(x^2-16)+x|)/4))+C