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

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

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

- ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

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

= - ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

Add a constant to the solution

= - ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

f (x) = 3 x^{4} + 2 x - 1, (2, 51)

\int (tan (x))^{7} (sec (x))^{2} d x

x \to 0 + lim (\frac{1}{x} \cdot sin (\frac{1}{x}))

\frac{\partial}{\partial x} (4 ln (x y))

x^{3} - 6 x^{2} - 34 x + 40, at x = 6

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