What is the integral of-(20arctan(2x))/(1+4x^2) ?

The integral of-(20arctan(2x))/(1+4x^2) is -5arctan^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-(20arctan(2x))/(1+4x^2)

\int - \frac{20 arctan ( 2 x )}{1 + 4 x ^{2}} d x

- 5 arctan^{2} (2 x) + C

Solution steps

\int - \frac{20 arctan ( 2 x )}{1 + 4 x ^{2}} d x

Take the constant out:

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

= - 20 \cdot \int \frac{arctan ( 2 x )}{1 + 4 x ^{2}} d x

Apply u-substitution

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

= - 20 \cdot \int \frac{arctan ( u )}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= - 20 \cdot \frac{1}{2} \cdot \int \frac{arctan ( u )}{1 + u ^{2}} d u

Apply u-substitution

: \int v d v

= - 20 \cdot \frac{1}{2} \cdot \int v d v

Apply the Power Rule

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

= - 20 \cdot \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

= - 20 \cdot \frac{1}{2} \cdot \frac{arctan ^{2} ( 2 x )}{2}

Simplify

- 20 \cdot \frac{1}{2} \cdot \frac{arctan ^{2} ( 2 x )}{2} : - 5 arctan^{2} (2 x)

= - 5 arctan^{2} (2 x)

Add a constant to the solution

= - 5 arctan^{2} (2 x) + C

\int (5 x^{2} + 7 x + \frac{1}{x ^{3}}) d x

\int x^{9} ln (3 x^{5}) d x

\int \frac{3 x + 7}{x ^{2} + 6 x + 25} d x

\int sin (2 + x) d x

\int sec^{1} (x) d x

What is the integral of-(20arctan(2x))/(1+4x^2) ?
The integral of-(20arctan(2x))/(1+4x^2) is -5arctan^2(2x)+C