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

The integral of 3/(1+9x^2) is arctan(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

arctan (3 x) + C

Solution steps

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

Apply Integral Substitution

= 3 \cdot \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 3 \cdot \frac{1}{3} arctan (u)

Substitute back

u = 3 x

= 3 \cdot \frac{1}{3} arctan (3 x)

Simplify

3 \cdot \frac{1}{3} arctan (3 x) : arctan (3 x)

= arctan (3 x)

Add a constant to the solution

= arctan (3 x) + C

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

\int x sin (3 x - 1) d x

\int \frac{tan ( x )}{ln ( cos ( x ) )} d x

\int (4 x + 3) (2 x^{2} + 3 x) 3 d x