What is the integral of (3x)/(1+9x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x)/(1+9x^2)

\int \frac{3 x}{1 + 9 x ^{2}} d x

\frac{1}{6} ln 1 + 9 x^{2} + C

Solution steps

\int \frac{3 x}{1 + 9 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{1 + 9 x ^{2}} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{18} ln ∣ u ∣

Substitute back

u = 1 + 9 x^{2}

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

Simplify

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

= \frac{1}{6} ln 1 + 9 x^{2}

Add a constant to the solution

= \frac{1}{6} ln 1 + 9 x^{2} + C

\int \frac{4}{3 x ^{2}} d x

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

\int 5 csc (x) cot (x) d x

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

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

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