What is the integral from 0 to 1 of 4arctan(-x) ?

The integral from 0 to 1 of 4arctan(-x) is 2ln(2)-pi

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integral from 0 to 1 of 4arctan(-x)

\int_{0}^{1} 4 arctan (- x) d x

2 ln (2) - π

Solution steps

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

Apply Integration By Parts

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

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

\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}

Simplify

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

Compute the boundaries:

\frac{1}{2} ln (2) - \frac{π}{4}

= 4 (\frac{1}{2} ln (2) - \frac{π}{4})

Simplify

= 2 ln (2) - π

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What is the integral from 0 to 1 of 4arctan(-x) ?
The integral from 0 to 1 of 4arctan(-x) is 2ln(2)-pi