What is the integral of (2x)/(x^2-x-6) ?

The integral of (2x)/(x^2-x-6) is ln|4x^2-4x-24|-1/5 (ln| 2/5 x+4/5 |-ln| 2/5 x-6/5 |)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(x^2-x-6)

\int \frac{2 x}{x ^{2} - x - 6} d x

ln 4 x^{2} - 4 x - 24 - \frac{1}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + C

Solution steps

\int \frac{2 x}{x ^{2} - x - 6} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x ^{2} - x - 6} d x

Complete the square

x^{2} - x - 6 : (x - \frac{1}{2})^{2} - \frac{25}{4}

= 2 \cdot \int \frac{x}{( x - \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

= 2 \cdot \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{2 u + 1}{4 u ^{2} - 25} d u

Expand

\frac{2 u + 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{1}{4 u ^{2} - 25}

Apply the Sum Rule:

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

= 2 \cdot 2 (\int \frac{2 u}{4 u ^{2} - 25} d u + \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 \cdot 2 (\frac{1}{4} ln 4 u^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

Substitute back

u = x - \frac{1}{2}

= 2 \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x - \frac{1}{2}) + 1 - ln \frac{2}{5} (x - \frac{1}{2}) - 1))

Simplify

2 \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x - \frac{1}{2}) + 1 - ln \frac{2}{5} (x - \frac{1}{2}) - 1)) : ln 4 x^{2} - 4 x - 24 - \frac{1}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5})

= ln 4 x^{2} - 4 x - 24 - \frac{1}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5})

Add a constant to the solution

= ln 4 x^{2} - 4 x - 24 - \frac{1}{5} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + C

\int 9 t e^{3 t} d t

\int \frac{2 x + 3}{x ^{2} + 3 x + 5} d x

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

\int \frac{x ^{5}}{( 1 + x ^{3} ) ^{3}} d x

What is the integral of (2x)/(x^2-x-6) ?
The integral of (2x)/(x^2-x-6) is ln|4x^2-4x-24|-1/5 (ln| 2/5 x+4/5 |-ln| 2/5 x-6/5 |)+C