What is the integral of (x^2+1)/(x^2+x-6) ?

The integral of (x^2+1)/(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 (x^2+1)/(x^2+x-6)

\int \frac{x ^{2} + 1}{x ^{2} + x - 6} d x

x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣ - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{2} + x - 6} d x

Expand

\frac{x ^{2} + 1}{x ^{2} + x - 6} : \frac{x ^{2}}{x ^{2} + x - 6} + \frac{1}{x ^{2} + x - 6}

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{x ^{2}}{x ^{2} + x - 6} d x + \int \frac{1}{x ^{2} + x - 6} d x

\int \frac{x ^{2}}{x ^{2} + x - 6} d x = x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣

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

= x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣ - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)

Add a constant to the solution

= x + \frac{4}{5} ln ∣ x - 2 ∣ - \frac{9}{5} ln ∣ x + 3 ∣ - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C

\int 2 sin (x) - 2 cos (2 x) d x

\int \frac{1}{x ( ln ( x ^{4} ) ) ^{8}} d x

\int (16 x + 1) 8 x^{2} + x d x

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

\int x^{6} e^{(x^{7} - 4)} d x

What is the integral of (x^2+1)/(x^2+x-6) ?
The integral of (x^2+1)/(x^2+x-6) is