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

The integral of (2x^2)/(6x^3+1) is 1/9 ln|6x^3+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{9} ln 6 x^{3} + 1 + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{2}}{6 x ^{3} + 1} d x

Apply u-substitution

: \int \frac{1}{18 u} d u

= 2 \cdot \int \frac{1}{18 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{18} ln ∣ u ∣

Substitute back

u = 6 x^{3} + 1

= 2 \cdot \frac{1}{18} ln 6 x^{3} + 1

Simplify

2 \cdot \frac{1}{18} ln 6 x^{3} + 1 : \frac{1}{9} ln 6 x^{3} + 1

= \frac{1}{9} ln 6 x^{3} + 1

Add a constant to the solution

= \frac{1}{9} ln 6 x^{3} + 1 + C

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

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

\int t^{3} e^{13 t} d t

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

What is the integral of (2x^2)/(6x^3+1) ?
The integral of (2x^2)/(6x^3+1) is 1/9 ln|6x^3+1|+C