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

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

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

\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{3}}{x - 1} d x

Apply u-substitution

: \int \frac{( u + 1 ) ^{3}}{u} d u

= \int \frac{( u + 1 ) ^{3}}{u} d u

Expand

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

= \int u^{2} + 3 u + 3 + \frac{1}{u} 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

= \int u^{2} d u + \int 3 u d u + \int 3 d u + \int \frac{1}{u} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣

Substitute back

u = x - 1

= \frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ + C

\frac{x}{1 + x}

\frac{\partial}{\partial x} (\frac{x}{y ^{3}} + \frac{3 y}{x ^{3}})

x^{2} - 5 x + 5 (0.5)

h \to 0 lim (\frac{( 3 ln ( e + h ) - 3 ln ( e ) )}{h})

\int \frac{4}{x ^{3} - 5 x ^{2} + 7 x - 3} d x

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