What is the integral of 1/(2xsqrt(16x^2-8)) ?

The integral of 1/(2xsqrt(16x^2-8)) is 1/(4sqrt(2))arcsec(sqrt(2)x)+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(16x^2-8))

\int \frac{1}{2 x 16 x ^{2} - 8} d x

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

Solution steps

\int \frac{1}{2 x 16 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 16 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{4}{2 2} x)

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

Simplify

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

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

Add a constant to the solution

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

\int 4^{3 x} d x

6 t^{2} + 8

\int \frac{14}{14 + e ^{x}} d x

f (t) = t^{3} - 2 t^{2} + 5 t + 4

f (x) = \frac{e ^{- x}}{x + 1}, at x = 1

What is the integral of 1/(2xsqrt(16x^2-8)) ?
The integral of 1/(2xsqrt(16x^2-8)) is 1/(4sqrt(2))arcsec(sqrt(2)x)+C