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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Apply Integral Reduction

= 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) ∣

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

Substitute back

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

= 2 \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

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

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

Add a constant to the solution

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

d er i v a t i v e y = \frac{3 x}{1 - x}

\frac{\partial}{\partial x} (cosh (y) + x - 2)

\int_{\frac{π}{4}}^{π} (sin (x)) d x

\int_{4}^{\infty} e^{5 x} d x

\int \frac{2 + x ^{2}}{1 + x ^{2}} d x

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