What is the integral of 1/(sqrt(x^2-4x)) ?

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

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integral of 1/(sqrt(x^2-4x))

\int \frac{1}{x ^{2} - 4 x} d x

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

Solution steps

\int \frac{1}{x ^{2} - 4 x} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 4} d u

= \int \frac{1}{u ^{2} - 4} 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}{2} (x - 2))) + sec (arcsec (\frac{1}{2} (x - 2)))

Simplify

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

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

Add a constant to the solution

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

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

\frac{d y}{d x} = \frac{3}{x y ^{2}}

\int \frac{cos ( 2 ln ( x ) ) ln ( x )}{x} d x

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

f (x) = 7 + 5 x

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