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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 8 x : (x - 4)^{2} - 16

= \int \frac{1}{( x - 4 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 16} d u

= \int \frac{1}{u ^{2} - 16} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arcsec (\frac{1}{4} (x - 4))) + sec (arcsec (\frac{1}{4} (x - 4)))

Simplify

ln tan (arcsec (\frac{1}{4} (x - 4))) + sec (arcsec (\frac{1}{4} (x - 4))) : ln (\frac{1}{4} x^{2} - 8 x + x - 4)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 y cos (x y))

x \to - 3 lim (- 4 x + 8)

\frac{\partial}{\partial x} (\frac{1}{ln ( x )})

\frac{\partial}{\partial x} (e^{x^{2} + y^{2} + z^{2}})

\frac{\partial}{\partial x} (\frac{x}{x + y})

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