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

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

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

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

\frac{1}{2} x x^{2} - a^{2} + \frac{a ^{2}}{2} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + C

Solution steps

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

Apply Trigonometric Substitution

= \int a^{2} sec^{3} (u) d u

Take the constant out:

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

= a^{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= a^{2} (\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) ∣

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

Substitute back

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

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

Simplify

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

= \frac{1}{2} x x^{2} - a^{2} + \frac{a ^{2}}{2} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣})

Add a constant to the solution

= \frac{1}{2} x x^{2} - a^{2} + \frac{a ^{2}}{2} ln (\frac{x ^{2} - a ^{2} + x}{∣ a ∣}) + C

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