What is the integral from 1 to sqrt(3 of) 6/(9+9x^2) ?

The integral from 1 to sqrt(3 of) 6/(9+9x^2) is pi/(18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to sqrt(3 of) 6/(9+9x^2)

\int_{1}^{3} \frac{6}{9 + 9 x ^{2}} d x

\frac{π}{18}

Decimal Notation

0.17453 \dots

Solution steps

\int_{1}^{3} \frac{6}{9 + 9 x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{3} \frac{1}{9 + 9 x ^{2}} d x

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

= 6 \cdot \int_{1}^{3} \frac{1}{9} \cdot \frac{1}{1 + x ^{2}} d x

Take the constant out:

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

= 6 \cdot \frac{1}{9} \cdot \int_{1}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 6 \cdot \frac{1}{9} [arctan (x)]_{1}^{3}

Simplify

6 \cdot \frac{1}{9} [arctan (x)]_{1}^{3} : \frac{2}{3} [arctan (x)]_{1}^{3}

= \frac{2}{3} [arctan (x)]_{1}^{3}

Compute the boundaries:

\frac{π}{12}

= \frac{2}{3} \cdot \frac{π}{12}

Simplify

= \frac{π}{18}

\int_{x^{2}}^{3} sin^{4} (t) d t

\int_{4}^{6} 5 x d x

\int_{0}^{1} (1 + x^{2}) d x

\int_{2}^{18} (4 x^{3} - 2 x + 2) d x

\int_{0}^{2} (x^{3} - x) d x

What is the integral from 1 to sqrt(3 of) 6/(9+9x^2) ?
The integral from 1 to sqrt(3 of) 6/(9+9x^2) is pi/(18)