What is the integral of 1/(sqrt(x^2-9x)) ?

The integral of 1/(sqrt(x^2-9x)) is ln(1/9 |2sqrt(x(x-9))+2x-9|)+C

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integral of 1/(sqrt(x^2-9x))

\int \frac{1}{x ^{2} - 9 x} d x

ln (\frac{1}{9} 2 x (x - 9) + 2 x - 9) + C

Solution steps

\int \frac{1}{x ^{2} - 9 x} d x

Complete the square

x^{2} - 9 x : (x - \frac{9}{2})^{2} - \frac{81}{4}

= \int \frac{1}{( x - \frac{9}{2} ) ^{2} - \frac{81}{4}} d x

Apply u-substitution

: \int \frac{2}{4 u ^{2} - 81} d u

= \int \frac{2}{4 u ^{2} - 81} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 u ^{2} - 81} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= 2 \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{9} (x - \frac{9}{2}))) + sec (arcsec (\frac{2}{9} (x - \frac{9}{2})))

Simplify

2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{9} (x - \frac{9}{2}))) + sec (arcsec (\frac{2}{9} (x - \frac{9}{2}))) : ln (\frac{1}{9} 2 x (x - 9) + 2 x - 9)

= ln (\frac{1}{9} 2 x (x - 9) + 2 x - 9)

Add a constant to the solution

= ln (\frac{1}{9} 2 x (x - 9) + 2 x - 9) + C

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What is the integral of 1/(sqrt(x^2-9x)) ?
The integral of 1/(sqrt(x^2-9x)) is ln(1/9 |2sqrt(x(x-9))+2x-9|)+C