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

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

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

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

Solution steps

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

Simplify

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

= \int \frac{x}{2 x + 1} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{2} d u

= \int \frac{u ^{2} - 1}{2} d u

Take the constant out:

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

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

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

\int 1 d u = u

= \frac{1}{2} (\frac{u ^{3}}{3} - u)

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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

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

\int e^{- x} arccot (e^{x}) d x

\int 5 x^{3} e^{2 x} d x

\int (2 + x)^{2} d x

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