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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

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

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

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{2}{3} x

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

Simplify

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

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

Add a constant to the solution

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

\int_{- \infty}^{- 1} e^{- 22 t} d t

f (x) = \frac{12}{x ^{2} - 4}, (4, 4)

\int 7 y d y

\frac{2 x \cdot 3 x - 1}{5 x}

\int \frac{e ^{x}}{e ^{2 x} + 16} d x

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