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

The integral of x/(sqrt(x^2-6)) is sqrt(x^2-6)+C

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

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

x^{2} - 6 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 6

= \frac{1}{2} \cdot 2 (x^{2} - 6)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (x^{2} - 6)^{\frac{1}{2}} : x^{2} - 6

= x^{2} - 6

Add a constant to the solution

= x^{2} - 6 + C

\frac{d}{d x} (\frac{4 - 3 x - x ^{2}}{x ^{2} - 1})

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

\frac{d h}{d t} = 5 h

1 + x^{2} d y - 1 - y^{2} d x = 0

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