What is the integral of (sqrt(2x^2-16))/(3x) ?

The integral of (sqrt(2x^2-16))/(3x) is 1/3 (sqrt(2x^2-16)-4arctan(1/4 sqrt(2x^2-16)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (sqrt(2x^2-16))/(3x)

\int \frac{2 x ^{2} - 16}{3 x} d x

\frac{1}{3} (2 x^{2} - 16 - 4 arctan (\frac{1}{4} 2 x^{2} - 16)) + C

Solution steps

\int \frac{2 x ^{2} - 16}{3 x} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{2 x ^{2} - 16}{x} d x

Apply Integration By Parts

= \frac{1}{3} (2 x^{2} - 16 - \int \frac{16}{x 2 x ^{2} - 16} d x)

\int \frac{16}{x 2 x ^{2} - 16} d x = 4 arctan (\frac{1}{4} 2 x^{2} - 16)

= \frac{1}{3} (2 x^{2} - 16 - 4 arctan (\frac{1}{4} 2 x^{2} - 16))

Add a constant to the solution

= \frac{1}{3} (2 x^{2} - 16 - 4 arctan (\frac{1}{4} 2 x^{2} - 16)) + C

What is the integral of (sqrt(2x^2-16))/(3x) ?
The integral of (sqrt(2x^2-16))/(3x) is 1/3 (sqrt(2x^2-16)-4arctan(1/4 sqrt(2x^2-16)))+C