What is the integral of (15x^2)/(sqrt(x^2-4)) ?

The integral of (15x^2)/(sqrt(x^2-4)) is 60(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C

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integral of (15x^2)/(sqrt(x^2-4))

\int \frac{15 x ^{2}}{x ^{2} - 4} d x

60 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

Solution steps

\int \frac{15 x ^{2}}{x ^{2} - 4} d x

Take the constant out:

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

= 15 \cdot \int \frac{x ^{2}}{x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int 4 sec^{3} (u) d u

= 15 \cdot \int 4 sec^{3} (u) d u

Take the constant out:

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

= 15 \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 15 \cdot 4 (\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) ∣

= 15 \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{2} x)

= 15 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)))

Simplify

15 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : 60 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

= 60 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

Add a constant to the solution

= 60 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

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What is the integral of (15x^2)/(sqrt(x^2-4)) ?
The integral of (15x^2)/(sqrt(x^2-4)) is 60(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C