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

The integral of (arctan^3(3x))/(1+9x^2) is 1/12 arctan^4(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (arctan^3(3x))/(1+9x^2)

\int \frac{arctan ^{3} ( 3 x )}{1 + 9 x ^{2}} d x

\frac{1}{12} arctan^{4} (3 x) + C

Solution steps

\int \frac{arctan ^{3} ( 3 x )}{1 + 9 x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{arctan ^{3} ( u )}{1 + u ^{2}} d u

Apply u-substitution

: \int v^{3} d v

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

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{3} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{3} \cdot \frac{arctan ^{4} ( 3 x )}{4}

Simplify

\frac{1}{3} \cdot \frac{arctan ^{4} ( 3 x )}{4} : \frac{1}{12} arctan^{4} (3 x)

= \frac{1}{12} arctan^{4} (3 x)

Add a constant to the solution

= \frac{1}{12} arctan^{4} (3 x) + C

\int x^{n} cos (a) x d x

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

\int \frac{sin ( x )}{1 + 2 cos ^{2} ( x )} d x

\int tan (x) (tan (x) + 1) d x

\int \frac{sec ^{2} ( 6 x )}{3 tan ( 6 x ) + 15} d x

What is the integral of (arctan^3(3x))/(1+9x^2) ?
The integral of (arctan^3(3x))/(1+9x^2) is 1/12 arctan^4(3x)+C