What is the integral of ((x^3-1))/(x^2(x-2)^3) ?

The integral of ((x^3-1))/(x^2(x-2)^3) is 3/16 ln|x|-1/(8x)-3/16 ln|x-2|-5/(4(x-2))-7/(8(x-2)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^3-1))/(x^2(x-2)^3)

\int \frac{( x ^{3} - 1 )}{x ^{2} ( x - 2 ) ^{3}} d x

\frac{3}{16} ln ∣ x ∣ - \frac{1}{8 x} - \frac{3}{16} ln ∣ x - 2 ∣ - \frac{5}{4 ( x - 2 )} - \frac{7}{8 ( x - 2 ) ^{2}} + C

Solution steps

\int \frac{x ^{3} - 1}{x ^{2} ( x - 2 ) ^{3}} d x

Take the partial fraction of

\frac{x ^{3} - 1}{x ^{2} ( x - 2 ) ^{3}} : \frac{3}{16 x} + \frac{1}{8 x ^{2}} - \frac{3}{16 ( x - 2 )} + \frac{5}{4 ( x - 2 ) ^{2}} + \frac{7}{4 ( x - 2 ) ^{3}}

= \int \frac{3}{16 x} + \frac{1}{8 x ^{2}} - \frac{3}{16 ( x - 2 )} + \frac{5}{4 ( x - 2 ) ^{2}} + \frac{7}{4 ( x - 2 ) ^{3}} 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{3}{16 x} d x + \int \frac{1}{8 x ^{2}} d x - \int \frac{3}{16 ( x - 2 )} d x + \int \frac{5}{4 ( x - 2 ) ^{2}} d x + \int \frac{7}{4 ( x - 2 ) ^{3}} d x

\int \frac{3}{16 x} d x = \frac{3}{16} ln ∣ x ∣

\int \frac{1}{8 x ^{2}} d x = - \frac{1}{8 x}

\int \frac{3}{16 ( x - 2 )} d x = \frac{3}{16} ln ∣ x - 2 ∣

\int \frac{5}{4 ( x - 2 ) ^{2}} d x = - \frac{5}{4 ( x - 2 )}

\int \frac{7}{4 ( x - 2 ) ^{3}} d x = - \frac{7}{8 ( x - 2 ) ^{2}}

= \frac{3}{16} ln ∣ x ∣ - \frac{1}{8 x} - \frac{3}{16} ln ∣ x - 2 ∣ - \frac{5}{4 ( x - 2 )} - \frac{7}{8 ( x - 2 ) ^{2}}

Add a constant to the solution

= \frac{3}{16} ln ∣ x ∣ - \frac{1}{8 x} - \frac{3}{16} ln ∣ x - 2 ∣ - \frac{5}{4 ( x - 2 )} - \frac{7}{8 ( x - 2 ) ^{2}} + C

\frac{\partial}{\partial x} (e^{1 + y - x})

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

y^{^{'}} = y - t, y (0) = 0.5

\int (\frac{4}{x ^{3}}) d x

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

What is the integral of ((x^3-1))/(x^2(x-2)^3) ?
The integral of ((x^3-1))/(x^2(x-2)^3) is 3/16 ln|x|-1/(8x)-3/16 ln|x-2|-5/(4(x-2))-7/(8(x-2)^2)+C