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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 1 + 3 x^{2}

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

Add a constant to the solution

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

\int (3 e^{x} + 8 x^{3} + 4 x) d x

\frac{d}{d x} (ln (x + x^{2} - 5))

x \to 0 - lim (h (x))

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