What is the integral of (x^2)/(sqrt(x^2-3)) ?

The integral of (x^2)/(sqrt(x^2-3)) is 3(1/6 xsqrt(x^2-3)+1/2 ln|sqrt(1/3 (x^2-3))+1/(sqrt(3))x|)+C

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integral of (x^2)/(sqrt(x^2-3))

\int \frac{x ^{2}}{x ^{2} - 3} d x

3 (\frac{1}{6} x x^{2} - 3 + \frac{1}{2} ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 3} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= 3 (\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) ∣

= 3 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= 3 \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

3 \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)) : 3 (\frac{1}{6} x x^{2} - 3 + \frac{1}{2} ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x)

= 3 (\frac{1}{6} x x^{2} - 3 + \frac{1}{2} ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x)

Add a constant to the solution

= 3 (\frac{1}{6} x x^{2} - 3 + \frac{1}{2} ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x) + C

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What is the integral of (x^2)/(sqrt(x^2-3)) ?
The integral of (x^2)/(sqrt(x^2-3)) is 3(1/6 xsqrt(x^2-3)+1/2 ln|sqrt(1/3 (x^2-3))+1/(sqrt(3))x|)+C