What is the integral of 9/(x^2+16) ?

The integral of 9/(x^2+16) is 9/4 arctan(x/4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(x^2+16)

\int \frac{9}{x ^{2} + 16} d x

\frac{9}{4} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{9}{x ^{2} + 16} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Integral Substitution

= 9 \cdot \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

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

= 9 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= 9 \cdot \frac{1}{4} arctan (\frac{x}{4})

Simplify

9 \cdot \frac{1}{4} arctan (\frac{x}{4}) : \frac{9}{4} arctan (\frac{x}{4})

= \frac{9}{4} arctan (\frac{x}{4})

Add a constant to the solution

= \frac{9}{4} arctan (\frac{x}{4}) + C

\int \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )} d x

\frac{x ^{2}}{3 + x}

ln (1 + x), 3

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

x \frac{d y}{d x} = y (ln (y) - ln (x))

What is the integral of 9/(x^2+16) ?
The integral of 9/(x^2+16) is 9/4 arctan(x/4)+C