What is the integral of ((x-2))/((2x^2-8x-1)) ?

The integral of ((x-2))/((2x^2-8x-1)) is 1/4 ln|2x^2-8x-1|+C

integral of ((x-2))/((2x^2-8x-1))

integral

\frac{( x - 2 )}{( 2 x ^{2} - 8 x - 1 )}

\frac{1}{4} ln 2 x^{2} - 8 x - 1 + C

Solution steps

\int \frac{x - 2}{2 x ^{2} - 8 x - 1} d x

Apply u-substitution

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} - 8 x - 1

= \frac{1}{4} ln 2 x^{2} - 8 x - 1

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} - 8 x - 1 + C

in t e g r a l \frac{x}{x - 6}

in t e g r a l (1 + x) cos (1 + x)

in t e g r a l \frac{( - e ^{- 2 x} )}{2}

in t e g r a l \frac{lo g _{10} ( 1 )}{x}

in t e g r a l \frac{x}{( x ^{2} + 1 ) ^{3}}

What is the integral of ((x-2))/((2x^2-8x-1)) ?
The integral of ((x-2))/((2x^2-8x-1)) is 1/4 ln|2x^2-8x-1|+C