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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/(x^2+9)

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

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

Solution steps

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

Apply Integral Substitution

= 5 \cdot \int \frac{1}{3 ( 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}{3} \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}{3} arctan (u)

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{sin ( x ^{2} )} d x

y = x^{2} - 2 x + 1, (- 1, 4)

\frac{\partial}{\partial x} (5^{8^{x^{2}}})

\int (e^{x} + 1) \cdot (x - 4) d x

\frac{d y}{d x} - \frac{1}{x} y = 1 - \frac{1}{x}

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