What is the integral of (arctan(8x))/(1+64x^2) ?

The integral of (arctan(8x))/(1+64x^2) is 1/16 arctan^2(8x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (arctan(8x))/(1+64x^2)

\int \frac{arctan ( 8 x )}{1 + 64 x ^{2}} d x

\frac{1}{16} arctan^{2} (8 x) + C

Solution steps

\int \frac{arctan ( 8 x )}{1 + 64 x ^{2}} d x

Apply Integral Substitution

= \int \frac{arctan ( u )}{8 ( 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}{8} \cdot \int \frac{arctan ( u )}{1 + u ^{2}} d u

Apply u-substitution

= \frac{1}{8} \cdot \int v d v

Apply the Power Rule

= \frac{1}{8} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{8} \cdot \frac{arctan ^{2} ( 8 x )}{2}

Simplify

\frac{1}{8} \cdot \frac{arctan ^{2} ( 8 x )}{2} : \frac{1}{16} arctan^{2} (8 x)

= \frac{1}{16} arctan^{2} (8 x)

Add a constant to the solution

= \frac{1}{16} arctan^{2} (8 x) + C

\int \frac{3 x}{x ^{2} - 5} d x

\int \frac{x ^{2} + 2 x + 3}{x ^{2} - 3 x + 2} d x

\int \frac{x e ^{2 y}}{4 - y} d x

\int \frac{10}{5 ^{3 x}} d x

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

What is the integral of (arctan(8x))/(1+64x^2) ?
The integral of (arctan(8x))/(1+64x^2) is 1/16 arctan^2(8x)+C