What is the integral of (x^2)/(sqrt(x^2-6)) ?

The integral of (x^2)/(sqrt(x^2-6)) is 6(1/12 xsqrt(x^2-6)+1/2 ln|sqrt(1/6 (x^2-6))+1/(sqrt(6))x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(x^2-6))

\int \frac{x ^{2}}{x ^{2} - 6} d x

6 (\frac{1}{12} x x^{2} - 6 + \frac{1}{2} ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 6} d x

Apply Trigonometric Substitution

: \int 6 sec^{3} (u) d u

= \int 6 sec^{3} (u) d u

Take the constant out:

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

= 6 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 6 (\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) ∣

= 6 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{6} x)

= 6 \frac{sec ^{2} ( arcsec ( \frac{1}{6} x ) ) sin ( arcsec ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x))

Simplify

6 \frac{sec ^{2} ( arcsec ( \frac{1}{6} x ) ) sin ( arcsec ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) : 6 (\frac{1}{12} x x^{2} - 6 + \frac{1}{2} ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x)

= 6 (\frac{1}{12} x x^{2} - 6 + \frac{1}{2} ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x)

Add a constant to the solution

= 6 (\frac{1}{12} x x^{2} - 6 + \frac{1}{2} ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x) + C

\frac{\partial}{\partial x} (5 e^{x y})

\frac{d}{d x} (\frac{3 x - 5}{x ^{2} + 7})

\frac{d}{d x} (- 12 e^{- 2 x})

x \to 3 lim (\frac{3 x - 9}{x - 3})

\frac{d}{d x} (\frac{tan ( x )}{x ^{3}})

What is the integral of (x^2)/(sqrt(x^2-6)) ?
The integral of (x^2)/(sqrt(x^2-6)) is 6(1/12 xsqrt(x^2-6)+1/2 ln|sqrt(1/6 (x^2-6))+1/(sqrt(6))x|)+C