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

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

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

ln (\frac{x ^{2} - 2 a x + x - a}{∣ a ∣}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Simplify

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

= ln (\frac{x ^{2} - 2 a x + x - a}{∣ a ∣})

Add a constant to the solution

= ln (\frac{x ^{2} - 2 a x + x - a}{∣ a ∣}) + C

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

\int \frac{ln ( 2 x + 1 )}{x} d x

\int sin (7 t) sec^{2} (cos (7 t)) d t

\int \frac{x - 1}{( x - 2 ) ( x + 3 )} d x

\int e^{x} + 20 d x

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