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

The integral of ((x-3))/(x^3+x^2) is 4ln|x|+3/x-4ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{( x - 3 )}{x ^{3} + x ^{2}} d x

4 ln ∣ x ∣ + \frac{3}{x} - 4 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

= 4 ln ∣ x ∣ - (- \frac{3}{x}) - 4 ln ∣ x + 1 ∣

Simplify

= 4 ln ∣ x ∣ + \frac{3}{x} - 4 ln ∣ x + 1 ∣

Add a constant to the solution

= 4 ln ∣ x ∣ + \frac{3}{x} - 4 ln ∣ x + 1 ∣ + C

\int 4 x^{2} - sin (x) d x

\int \frac{1}{x ^{2} - 36} d x

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

y^{^{'}} = - a \cdot y^{2}

\frac{d}{d t} (t^{9} e^{t})

What is the integral of ((x-3))/(x^3+x^2) ?
The integral of ((x-3))/(x^3+x^2) is 4ln|x|+3/x-4ln|x+1|+C