What is the integral of (3x^2+4x+1)/(3x^3+6x^2+3x+5) ?

The integral of (3x^2+4x+1)/(3x^3+6x^2+3x+5) is 1/3 ln|3x^3+6x^2+3x+5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2+4x+1)/(3x^3+6x^2+3x+5)

\int \frac{3 x ^{2} + 4 x + 1}{3 x ^{3} + 6 x ^{2} + 3 x + 5} d x

\frac{1}{3} ln 3 x^{3} + 6 x^{2} + 3 x + 5 + C

Solution steps

\int \frac{3 x ^{2} + 4 x + 1}{3 x ^{3} + 6 x ^{2} + 3 x + 5} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x^{3} + 6 x^{2} + 3 x + 5

= \frac{1}{3} ln 3 x^{3} + 6 x^{2} + 3 x + 5

Add a constant to the solution

= \frac{1}{3} ln 3 x^{3} + 6 x^{2} + 3 x + 5 + C

y = x^{2} sin^{6} (x) + x cos^{- 4} (x)

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

\frac{d}{d x} (\frac{3 x ^{4}}{4})

\int \frac{2 x - 1}{x ^{2} - x - 12} d x

y^{^{''}} + 9 y = 5 e^{x} - 162 x

What is the integral of (3x^2+4x+1)/(3x^3+6x^2+3x+5) ?
The integral of (3x^2+4x+1)/(3x^3+6x^2+3x+5) is 1/3 ln|3x^3+6x^2+3x+5|+C