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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(1+9u^2)

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

3 arctan (3 u) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 9 \cdot \int \frac{1}{3 ( 1 + v ^{2} )} d v

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= 9 \cdot \frac{1}{3} arctan (v)

Substitute back

v = 3 u

= 9 \cdot \frac{1}{3} arctan (3 u)

Simplify

9 \cdot \frac{1}{3} arctan (3 u) : 3 arctan (3 u)

= 3 arctan (3 u)

Add a constant to the solution

= 3 arctan (3 u) + C

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

\int 6 - \frac{6}{x ^{2}} d x

\int e^{- (3 x^{2} + 1)} x d x

\int x cos (\frac{nπ x}{20}) d x

\int \frac{sec ^{2} ( x )}{5 + 3 tan ( x )} d x

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