What is the integral of 1/(sqrt(x^2-6x)) ?

The integral of 1/(sqrt(x^2-6x)) is ln(1/3 |sqrt(x^2-6x)+x-3|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x^2-6x))

\int \frac{1}{x ^{2} - 6 x} d x

ln (\frac{1}{3} x^{2} - 6 x + x - 3) + C

Solution steps

\int \frac{1}{x ^{2} - 6 x} d x

Complete the square

x^{2} - 6 x : (x - 3)^{2} - 9

= \int \frac{1}{( x - 3 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 9} d u

= \int \frac{1}{u ^{2} - 9} 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}{3} (x - 3))) + sec (arcsec (\frac{1}{3} (x - 3)))

Simplify

ln tan (arcsec (\frac{1}{3} (x - 3))) + sec (arcsec (\frac{1}{3} (x - 3))) : ln (\frac{1}{3} x^{2} - 6 x + x - 3)

= ln (\frac{1}{3} x^{2} - 6 x + x - 3)

Add a constant to the solution

= ln (\frac{1}{3} x^{2} - 6 x + x - 3) + C

n = 1 \sum \infty tan (3 n)

g (x) = \frac{x ^{2} - 64}{x + 8}

y^{^{''}} + 6 y^{^{'}} + 5 y = - 3 t e^{(4 t)}

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

What is the integral of 1/(sqrt(x^2-6x)) ?
The integral of 1/(sqrt(x^2-6x)) is ln(1/3 |sqrt(x^2-6x)+x-3|)+C