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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

2 ln \frac{x - a}{a} + \frac{x}{a} + C

Solution steps

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

Factor

\frac{1}{x ^{2} - a x}

= \int \frac{1}{x ( x - a )} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (x - a) = x (x - a)

= \int \frac{1}{x x - a} d x

Apply u-substitution

= \int \frac{2}{u ^{2} + a} 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}{u ^{2} + a} d u

Apply Trigonometric Substitution

= 2 \cdot \int sec (v) d v

Use the common integral:

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

= 2 ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 ln tan (arctan (\frac{1}{a} x - a)) + sec (arctan (\frac{1}{a} x - a))

Simplify

2 ln tan (arctan (\frac{1}{a} x - a)) + sec (arctan (\frac{1}{a} x - a)) : 2 ln \frac{x - a}{a} + \frac{x}{a}

= 2 ln \frac{x - a}{a} + \frac{x}{a}

Add a constant to the solution

= 2 ln \frac{x - a}{a} + \frac{x}{a} + C

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