What is the integral from 0 to 1.5 of 4arctan(x) ?

The integral from 0 to 1.5 of 4arctan(x) is 3.53945…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1.5 of 4arctan(x)

\int_{0}^{1.5} 4 arctan (x) d x

3.53945 \dots

Solution steps

\int_{0}^{1.5} 4 arctan (x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1.5} arctan (x) d x

Apply Integration By Parts

: [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{0}^{1.5}

= 4 [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{0}^{1.5}

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= 4 [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{1.5}

Simplify

4 [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{1.5} : 4 [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{1.5}

= 4 [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{1.5}

Compute the boundaries:

0.88486 \dots

= 4 \cdot 0.88486 \dots

Simplify

= 3.53945 \dots

\int_{0}^{x} \frac{1}{( 1 - x ) ^{2} ( 1 - x )} d x

\int_{- B}^{B} (- B^{2} - x^{2}) d x

\int_{1}^{2} (4 - x) d x

\int_{0}^{t^{2}} u^{7} d u

\int_{1}^{2} \frac{( x ^{2} + 2 x + 4 )}{x} d x

What is the integral from 0 to 1.5 of 4arctan(x) ?
The integral from 0 to 1.5 of 4arctan(x) is 3.53945…