What is the integral of (3x^2-8x)/(x^3-4x^2+1) ?

The integral of (3x^2-8x)/(x^3-4x^2+1) is ln|x^3-4x^2+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^2-8x)/(x^3-4x^2+1)

\int \frac{3 x ^{2} - 8 x}{x ^{3} - 4 x ^{2} + 1} d x

ln x^{3} - 4 x^{2} + 1 + C

Solution steps

\int \frac{3 x ^{2} - 8 x}{x ^{3} - 4 x ^{2} + 1} d x

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{3} - 4 x^{2} + 1

= ln x^{3} - 4 x^{2} + 1

Add a constant to the solution

= ln x^{3} - 4 x^{2} + 1 + C

\int x e^{\frac{- x ^{2}}{a}} d x

\int y \cdot ln (y) d y

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

\int 1 + cos (3 x) d x

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

What is the integral of (3x^2-8x)/(x^3-4x^2+1) ?
The integral of (3x^2-8x)/(x^3-4x^2+1) is ln|x^3-4x^2+1|+C