What is the integral of 5/(4+9r^2) ?

The integral of 5/(4+9r^2) is 5/6 arctan(3/2 r)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/(4+9r^2)

\int \frac{5}{4 + 9 r ^{2}} d r

\frac{5}{6} arctan (\frac{3}{2} r) + C

Solution steps

\int \frac{5}{4 + 9 r ^{2}} d r

Take the constant out:

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

= 5 \cdot \int \frac{1}{4 + 9 r ^{2}} d r

Apply Integral Substitution

= 5 \cdot \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{3}{2} r

= 5 \cdot \frac{1}{6} arctan (\frac{3}{2} r)

Simplify

5 \cdot \frac{1}{6} arctan (\frac{3}{2} r) : \frac{5}{6} arctan (\frac{3}{2} r)

= \frac{5}{6} arctan (\frac{3}{2} r)

Add a constant to the solution

= \frac{5}{6} arctan (\frac{3}{2} r) + C

x \to \infty lim (4 + x)

\int \frac{( x + 3 )}{( x + 4 )} d x

\frac{d}{d x} (4 x^{3} + 3 x^{2} + 2 x + 1)

y = 2 x^{2} - 5

n = 1 \sum \infty 5^{\frac{1}{n}}

What is the integral of 5/(4+9r^2) ?
The integral of 5/(4+9r^2) is 5/6 arctan(3/2 r)+C