What is the integral of (3x-1)/(2x^2+8) ?

The integral of (3x-1)/(2x^2+8) is 3/4 ln|2x^2+8|-1/4 arctan(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x-1)/(2x^2+8)

\int \frac{3 x - 1}{2 x ^{2} + 8} d x

\frac{3}{4} ln 2 x^{2} + 8 - \frac{1}{4} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{3 x - 1}{2 x ^{2} + 8} d x

Expand

\frac{3 x - 1}{2 x ^{2} + 8} : \frac{3 x}{2 x ^{2} + 8} - \frac{1}{2 x ^{2} + 8}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 x}{2 x ^{2} + 8} d x - \int \frac{1}{2 x ^{2} + 8} d x

\int \frac{3 x}{2 x ^{2} + 8} d x = \frac{3}{4} ln 2 x^{2} + 8

\int \frac{1}{2 x ^{2} + 8} d x = \frac{1}{4} arctan (\frac{x}{2})

= \frac{3}{4} ln 2 x^{2} + 8 - \frac{1}{4} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{3}{4} ln 2 x^{2} + 8 - \frac{1}{4} arctan (\frac{x}{2}) + C

What is the integral of (3x-1)/(2x^2+8) ?
The integral of (3x-1)/(2x^2+8) is 3/4 ln|2x^2+8|-1/4 arctan(x/2)+C