What is the integral of 1/(sqrt(2as+s^2)) ?

The integral of 1/(sqrt(2as+s^2)) is ln((|sqrt(s^2+2as)+s+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(2as+s^2))

\int \frac{1}{2 a s + s ^{2}} d s

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

Solution steps

\int \frac{1}{2 a s + s ^{2}} d s

Complete the square

2 a s + s^{2} : (s + a)^{2} - a^{2}

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

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} (s + a))) + sec (arcsec (\frac{1}{a} (s + a)))

Simplify

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

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

Add a constant to the solution

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

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

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

\int e^{c o s (5 t)} sin (5 t) d t

\int \frac{5}{x ln ( 2 x )} d x

\int cos^{5} (3 x) sin^{2} (3 x) d x

What is the integral of 1/(sqrt(2as+s^2)) ?
The integral of 1/(sqrt(2as+s^2)) is ln((|sqrt(s^2+2as)+s+a|)/(|a|))+C