What is the integral of (2x-1)/(5x^2-x+2) ?

The integral of (2x-1)/(5x^2-x+2) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-1)/(5x^2-x+2)

\int \frac{2 x - 1}{5 x ^{2} - x + 2} d x

20 (2 (\frac{1}{200} ln 100 x^{2} - 20 x + 40 + \frac{1}{100 39} arctan (\frac{10 x - 1}{39})) - \frac{1}{10 39} arctan (\frac{10 x - 1}{39})) + C

Solution steps

\int \frac{2 x - 1}{5 x ^{2} - x + 2} d x

Complete the square

5 x^{2} - x + 2 : 5 (x - \frac{1}{10})^{2} + \frac{39}{20}

= \int \frac{2 x - 1}{5 ( x - \frac{1}{10} ) ^{2} + \frac{39}{20}} d x

Apply u-substitution

: \int \frac{20 ( 2 ( u + \frac{1}{10} ) - 1 )}{100 u ^{2} + 39} d u

= \int \frac{20 ( 2 ( u + \frac{1}{10} ) - 1 )}{100 u ^{2} + 39} d u

Take the constant out:

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

= 20 \cdot \int \frac{2 ( u + \frac{1}{10} ) - 1}{100 u ^{2} + 39} d u

Expand

\frac{2 ( u + \frac{1}{10} ) - 1}{100 u ^{2} + 39} : \frac{2 ( u + \frac{1}{10} )}{100 u ^{2} + 39} - \frac{1}{100 u ^{2} + 39}

Apply the Sum Rule:

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

= 20 (\int \frac{2 ( u + \frac{1}{10} )}{100 u ^{2} + 39} d u - \int \frac{1}{100 u ^{2} + 39} d u)

\int \frac{2 ( u + \frac{1}{10} )}{100 u ^{2} + 39} d u = 2 (\frac{1}{200} ln 100 u^{2} + 39 + \frac{1}{100 39} arctan (\frac{10}{39} u))

\int \frac{1}{100 u ^{2} + 39} d u = \frac{1}{10 39} arctan (\frac{10}{39} u)

= 20 (2 (\frac{1}{200} ln 100 u^{2} + 39 + \frac{1}{100 39} arctan (\frac{10}{39} u)) - \frac{1}{10 39} arctan (\frac{10}{39} u))

Substitute back

u = x - \frac{1}{10}

= 20 (2 (\frac{1}{200} ln 100 (x - \frac{1}{10})^{2} + 39 + \frac{1}{100 39} arctan (\frac{10}{39} (x - \frac{1}{10}))) - \frac{1}{10 39} arctan (\frac{10}{39} (x - \frac{1}{10})))

Simplify

20 (2 (\frac{1}{200} ln 100 (x - \frac{1}{10})^{2} + 39 + \frac{1}{100 39} arctan (\frac{10}{39} (x - \frac{1}{10}))) - \frac{1}{10 39} arctan (\frac{10}{39} (x - \frac{1}{10}))) : 20 (2 (\frac{1}{200} ln 100 x^{2} - 20 x + 40 + \frac{1}{100 39} arctan (\frac{10 x - 1}{39})) - \frac{1}{10 39} arctan (\frac{10 x - 1}{39}))

= 20 (2 (\frac{1}{200} ln 100 x^{2} - 20 x + 40 + \frac{1}{100 39} arctan (\frac{10 x - 1}{39})) - \frac{1}{10 39} arctan (\frac{10 x - 1}{39}))

Add a constant to the solution

= 20 (2 (\frac{1}{200} ln 100 x^{2} - 20 x + 40 + \frac{1}{100 39} arctan (\frac{10 x - 1}{39})) - \frac{1}{10 39} arctan (\frac{10 x - 1}{39})) + C

\int \frac{2 x + 2}{x ^{2} + 2 x + 2} d x

\int x^{2} + 6 x + 25 d x

\int \frac{( y ^{4} - 8 )}{y ^{3} + 2 y ^{2}} d y

\int csc (1 - 2 x) cot (1 - 2 x) d x

\int \frac{b x + c}{x ^{2} - 8 x + 25} d x

What is the integral of (2x-1)/(5x^2-x+2) ?
The integral of (2x-1)/(5x^2-x+2) is