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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u + 1 ) ^{2}}{8 u} d u

= \int \frac{( u + 1 ) ^{2}}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{( u + 1 ) ^{2}}{u} d u

Expand

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

= \frac{1}{8} \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

= \frac{1}{8} (\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 ∣

= \frac{1}{8} (\frac{u ^{2}}{2} + 2 u + ln ∣ u ∣)

Substitute back

u = 2 x - 1

= \frac{1}{8} (\frac{( 2 x - 1 ) ^{2}}{2} + 2 (2 x - 1) + ln ∣ 2 x - 1 ∣)

Simplify

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

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

Add a constant to the solution

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

\int x (4 - x) d x

f (x) = (\frac{x}{1 + x})^{5}, at x = 1

8^{t} \cdot t + 10

\frac{d}{d y} (tan (x^{2} + y))

x^{3} - 4

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