What is the integral of 1/(sqrt(x^2+6x+8)) ?

The integral of 1/(sqrt(x^2+6x+8)) is ln|sqrt(x^2+6x+8)+x+3|+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+6x+8))

\int \frac{1}{x ^{2} + 6 x + 8} d x

ln x^{2} + 6 x + 8 + x + 3 + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 8} d x

Complete the square

x^{2} + 6 x + 8 : (x + 3)^{2} - 1

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

Apply u-substitution

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

= \int \frac{1}{u ^{2} - 1} 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 (x + 3)) + sec (arcsec (x + 3)) ∣

Simplify

ln ∣ tan (arcsec (x + 3)) + sec (arcsec (x + 3)) ∣ : ln x^{2} + 6 x + 8 + x + 3

= ln x^{2} + 6 x + 8 + x + 3

Add a constant to the solution

= ln x^{2} + 6 x + 8 + x + 3 + C

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

\int \frac{5}{4 x ^{2} + 32 x + 70} d x

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

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

\int e^{5 x} 5 d x

What is the integral of 1/(sqrt(x^2+6x+8)) ?
The integral of 1/(sqrt(x^2+6x+8)) is ln|sqrt(x^2+6x+8)+x+3|+C