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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2/(xsqrt(x^2-16))

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= 2 \cdot \int \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int 2^{x} \cdot e^{x} d x

\int sin^{3 x} (θ) d y

\int d arcsin (x) d x

\int \frac{5 - θ ^{2}}{θ ^{2}} d θ

\int arccos (5 x) d x

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