What is the integral of (4x^2)/(x-1) ?

The integral of (4x^2)/(x-1) is 4(((x-1)^2)/2+2(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 (4x^2)/(x-1)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

= 4 \cdot \int \frac{( u + 1 ) ^{2}}{u} d u

Expand

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

= 4 \cdot \int u + 2 + \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

= 4 (\int u d u + \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 4 (\frac{u ^{2}}{2} + 2 u + ln ∣ u ∣)

Substitute back

u = x - 1

= 4 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣)

Add a constant to the solution

= 4 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) + C

\int e^{3 x} cot (e^{3 x}) d x

\int \frac{( 2 x ^{2} - 5 x + 1 )}{( x + 2 ) ( x ^{2} + 8 )} d x

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

\int \frac{1}{x - 7} d x

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