What is the integral of (x^2)/(sqrt(x^2-5)) ?

The integral of (x^2)/(sqrt(x^2-5)) is 5(1/10 xsqrt(x^2-5)+1/2 ln|sqrt(1/5 (x^2-5))+1/(sqrt(5))x|)+C

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integral of (x^2)/(sqrt(x^2-5))

\int \frac{x ^{2}}{x ^{2} - 5} d x

5 (\frac{1}{10} x x^{2} - 5 + \frac{1}{2} ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 5} d x

Apply Trigonometric Substitution

: \int 5 sec^{3} (u) d u

= \int 5 sec^{3} (u) d u

Take the constant out:

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

= 5 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 5 (\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) ∣

= 5 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{5} x)

= 5 \frac{sec ^{2} ( arcsec ( \frac{1}{5} x ) ) sin ( arcsec ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))

Simplify

5 \frac{sec ^{2} ( arcsec ( \frac{1}{5} x ) ) sin ( arcsec ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) : 5 (\frac{1}{10} x x^{2} - 5 + \frac{1}{2} ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x)

= 5 (\frac{1}{10} x x^{2} - 5 + \frac{1}{2} ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x)

Add a constant to the solution

= 5 (\frac{1}{10} x x^{2} - 5 + \frac{1}{2} ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x) + C

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What is the integral of (x^2)/(sqrt(x^2-5)) ?
The integral of (x^2)/(sqrt(x^2-5)) is 5(1/10 xsqrt(x^2-5)+1/2 ln|sqrt(1/5 (x^2-5))+1/(sqrt(5))x|)+C