What is the integral of (2x^3+x^2-6x+7)/(x^2+x-6) ?

The integral of (2x^3+x^2-6x+7)/(x^2+x-6) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^3+x^2-6x+7)/(x^2+x-6)

\int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{x ^{2} + x - 6} d x

4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

Solution steps

\int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{x ^{2} + x - 6} d x

Complete the square

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

= \int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{4 ( - 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10 )}{4 u ^{2} - 25} d u

= \int \frac{4 ( - 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10 )}{4 u ^{2} - 25} d u

Take the constant out:

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

= 4 \cdot \int \frac{- 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10}{4 u ^{2} - 25} d u

Long division

\frac{- 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10}{4 u ^{2} - 25} : \frac{u}{2} - \frac{1}{2} + \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )}

= 4 \cdot \int \frac{u}{2} - \frac{1}{2} + \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} d u

Apply the Sum Rule:

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

= 4 (\int \frac{u}{2} d u - \int \frac{1}{2} d u + \int \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} d u)

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

\int \frac{1}{2} d u = \frac{1}{2} u

\int \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} d u = \frac{1}{2} (\frac{7}{4} ln 4 u^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

= 4 (\frac{u ^{2}}{4} - \frac{1}{2} u + \frac{1}{2} (\frac{7}{4} ln 4 u^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

Substitute back

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

= 4 (\frac{( x + \frac{1}{2} ) ^{2}}{4} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{2} (\frac{7}{4} ln 4 (x + \frac{1}{2})^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

Simplify

4 (\frac{( x + \frac{1}{2} ) ^{2}}{4} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{2} (\frac{7}{4} ln 4 (x + \frac{1}{2})^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

= 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

Add a constant to the solution

= 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

\int cos (y) d y

x \to 4 lim (\frac{1}{x - 3})

\frac{\partial}{\partial x} (\frac{x}{y} + \frac{1}{x} + \frac{1}{y})

\int_{- 1}^{3} x^{3} d x

x \to 2 lim (6^{x} + 2)

What is the integral of (2x^3+x^2-6x+7)/(x^2+x-6) ?
The integral of (2x^3+x^2-6x+7)/(x^2+x-6) is