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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/((1+4x^2)(arctan(2x))^3))

\int (\frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}}) d x

- \frac{1}{4 arctan ^{2} ( 2 x )} + C

Solution steps

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{3}} d v

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

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= \frac{1}{2} (- \frac{1}{2 v ^{2}})

Substitute back

= \frac{1}{2} (- \frac{1}{2 arctan ^{2} ( 2 x )})

Simplify

\frac{1}{2} (- \frac{1}{2 arctan ^{2} ( 2 x )}) : - \frac{1}{4 arctan ^{2} ( 2 x )}

= - \frac{1}{4 arctan ^{2} ( 2 x )}

Add a constant to the solution

= - \frac{1}{4 arctan ^{2} ( 2 x )} + C

\int \frac{( 2 x + 3 )}{( x + 7 ) ^{3}} d x

n = 1 \sum \infty 3 (\frac{1}{5})^{n - 1}

f (x) = e^{2 x} - 1

n = 1 \sum \infty - n

x \to 0 lim (\frac{sin ( 3 x ^{2} )}{ln ( x ^{2} + 1 )})

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