What is the integral from 1 to 5 of (4x^3-1)/(2x-1) ?

The integral from 1 to 5 of (4x^3-1)/(2x-1) is 290/3-1/2 ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 5 of (4x^3-1)/(2x-1)

\int_{1}^{5} \frac{4 x ^{3} - 1}{2 x - 1} d x

\frac{290}{3} - \frac{1}{2} ln (3)

Decimal Notation

96.11736 \dots

Solution steps

\int_{1}^{5} \frac{4 x ^{3} - 1}{2 x - 1} d x

Long division

\frac{4 x ^{3} - 1}{2 x - 1} : 2 x^{2} + x + \frac{1}{2} - \frac{1}{2 ( 2 x - 1 )}

= \int_{1}^{5} 2 x^{2} + x + \frac{1}{2} - \frac{1}{2 ( 2 x - 1 )} d x

Apply the Sum Rule:

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

= \int_{1}^{5} 2 x^{2} d x + \int_{1}^{5} x d x + \int_{1}^{5} \frac{1}{2} d x - \int_{1}^{5} \frac{1}{2 ( 2 x - 1 )} d x

\int_{1}^{5} 2 x^{2} d x = \frac{248}{3}

\int_{1}^{5} x d x = 12

\int_{1}^{5} \frac{1}{2} d x = 2

\int_{1}^{5} \frac{1}{2 ( 2 x - 1 )} d x = \frac{1}{2} ln (3)

= \frac{248}{3} + 12 + 2 - \frac{1}{2} ln (3)

Simplify

\frac{248}{3} + 12 + 2 - \frac{1}{2} ln (3) : \frac{290}{3} - \frac{1}{2} ln (3)

= \frac{290}{3} - \frac{1}{2} ln (3)

\int_{y^{2} - 4}^{5} (x + 2 y) d x

\int_{0}^{1} 2 π x 4 x - x^{2} d x

\int_{0}^{2} (2 - t)^{2} t^{3} d t

\int_{0}^{2} - x^{2} + 4 x - (x^{2}) d x

\int_{4}^{5} \frac{1}{x ^{2} - 5 x + 6} d x

What is the integral from 1 to 5 of (4x^3-1)/(2x-1) ?
The integral from 1 to 5 of (4x^3-1)/(2x-1) is 290/3-1/2 ln(3)