What is the integral of 1/(sqrt(x^2-10x)) ?

The integral of 1/(sqrt(x^2-10x)) is ln(1/5 |sqrt(x^2-10x)+x-5|)+C

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integral of 1/(sqrt(x^2-10x))

\int \frac{1}{x ^{2} - 10 x} d x

ln (\frac{1}{5} x^{2} - 10 x + x - 5) + C

Solution steps

\int \frac{1}{x ^{2} - 10 x} d x

Complete the square

x^{2} - 10 x : (x - 5)^{2} - 25

= \int \frac{1}{( x - 5 ) ^{2} - 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 25} d u

= \int \frac{1}{u ^{2} - 25} 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}{5} (x - 5))) + sec (arcsec (\frac{1}{5} (x - 5)))

Simplify

ln tan (arcsec (\frac{1}{5} (x - 5))) + sec (arcsec (\frac{1}{5} (x - 5))) : ln (\frac{1}{5} x^{2} - 10 x + x - 5)

= ln (\frac{1}{5} x^{2} - 10 x + x - 5)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} - 10 x + x - 5) + C

y^{^{'}} = y^{2} + 3

f (x) = 2 x^{2} + 3 x - 2, at a = - 1

\int_{5}^{5.89} 0.75 (y - 5)^{2} + 5 d y

\frac{d}{d x} (- x e^{- 2 x y})

0.01 (0.2 t^{2} + 4 t + 64)^{\frac{2}{3}}

What is the integral of 1/(sqrt(x^2-10x)) ?
The integral of 1/(sqrt(x^2-10x)) is ln(1/5 |sqrt(x^2-10x)+x-5|)+C