What is the integral of ((3-x)/(sqrt(x^2-6x+2))) ?

The integral of ((3-x)/(sqrt(x^2-6x+2))) is -sqrt(x^2-6x+2)+C

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integral of ((3-x)/(sqrt(x^2-6x+2)))

\int (\frac{3 - x}{x ^{2} - 6 x + 2}) d x

- x^{2} - 6 x + 2 + C

Solution steps

\int \frac{3 - x}{x ^{2} - 6 x + 2} d x

Factor

3 - x : - (- 3 + x)

= \int \frac{- ( - 3 + x )}{x ^{2} - 6 x + 2} d x

Take the constant out:

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

= - \int \frac{- 3 + x}{x ^{2} - 6 x + 2} 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 x + 2

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

Simplify

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

= - x^{2} - 6 x + 2

Add a constant to the solution

= - x^{2} - 6 x + 2 + C

\frac{d}{d x} (cos (\frac{π x}{4}))

f (x) = x^{2}, at x = - 3

x \to 4 lim (\frac{x ^{q (x)} ( 4 - x )}{e ^{4 - x} - 1})

x \to 9 lim (\frac{7}{x - 9})

1 - 3 lo g_{5} (x)

What is the integral of ((3-x)/(sqrt(x^2-6x+2))) ?
The integral of ((3-x)/(sqrt(x^2-6x+2))) is -sqrt(x^2-6x+2)+C