What is the integral of 1/(2xsqrt(4x^2-8)) ?

The integral of 1/(2xsqrt(4x^2-8)) is 1/(4sqrt(2))arcsec(x/(sqrt(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/(2xsqrt(4x^2-8))

\int \frac{1}{2 x 4 x ^{2} - 8} d x

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

Solution steps

\int \frac{1}{2 x 4 x ^{2} - 8} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x 4 x ^{2} - 8} d x

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \int \frac{1}{2 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 \frac{1}{2 2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot u

Substitute back

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

= \frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot arcsec (\frac{2}{2 2} x)

Simplify

\frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot arcsec (\frac{2}{2 2} x) : \frac{1}{4 2} arcsec (\frac{x}{2})

= \frac{1}{4 2} arcsec (\frac{x}{2})

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{n x y})

\frac{d}{d x} (9 x^{2})

\int e^{i 6 x} cos (3 x) d x

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

\int \frac{1}{3 x + 6} d x

What is the integral of 1/(2xsqrt(4x^2-8)) ?
The integral of 1/(2xsqrt(4x^2-8)) is 1/(4sqrt(2))arcsec(x/(sqrt(2)))+C