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

The integral of 1/(sqrt(2x^2+8)) is 1/(sqrt(2))ln((|x+sqrt(4+x^2)|)/2)+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(2x^2+8))

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

\frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{2}{2 2} x)

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

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{2 2} x)) + sec (arctan (\frac{2}{2 2} x)) : \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2})

= \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2}) + C

\int \frac{y}{2 - y} d y

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

\int arccos^{2} (x) d x

\int - z sin (z) + cos (z) d z

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

What is the integral of 1/(sqrt(2x^2+8)) ?
The integral of 1/(sqrt(2x^2+8)) is 1/(sqrt(2))ln((|x+sqrt(4+x^2)|)/2)+C