What is the integral of 1/(9+u^2) ?

The integral of 1/(9+u^2) is 1/3 arctan(u/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(9+u^2)

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

\frac{1}{3} arctan (\frac{u}{3}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

v = \frac{u}{3}

= \frac{1}{3} arctan (\frac{u}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{u}{3}) + C

\int \frac{cos ^{4} ( x )}{csc ^{3} ( x )} d x

\int \frac{x + 5}{x - 5} d x

\int - sin (3 x) (5 + cos (3 x))^{4} d x

\int \frac{1}{x ^{10}} d x

\int - x^{6} d x

What is the integral of 1/(9+u^2) ?
The integral of 1/(9+u^2) is 1/3 arctan(u/3)+C