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

The integral from 0 to 1/4 of arctan(4x) is (pi-2ln(2))/(16)

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

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

\frac{π - 2 ln ( 2 )}{16}

Decimal

0.10970 \dots

Solution steps

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

Apply Integration By Parts

: [x arctan (4 x) - \int \frac{4 x}{16 x ^{2} + 1} d x]_{0}^{\frac{1}{4}}

= [x arctan (4 x) - \int \frac{4 x}{16 x ^{2} + 1} d x]_{0}^{\frac{1}{4}}

\int \frac{4 x}{16 x ^{2} + 1} d x = \frac{1}{8} ln 16 x^{2} + 1

= [x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1]_{0}^{\frac{1}{4}}

Simplify

[x arctan (4 x) - \frac{1}{8} ln 16 x^{2} + 1]_{0}^{\frac{1}{4}} : [\frac{1}{8} (8 x arctan (4 x) - ln 16 x^{2} + 1)]_{0}^{\frac{1}{4}}

= [\frac{1}{8} (8 x arctan (4 x) - ln 16 x^{2} + 1)]_{0}^{\frac{1}{4}}

Compute the boundaries:

\frac{π - 2 ln ( 2 )}{16}

= \frac{π - 2 ln ( 2 )}{16}

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What is the integral from 0 to 1/4 of arctan(4x) ?
The integral from 0 to 1/4 of arctan(4x) is (pi-2ln(2))/(16)