What is the integral from 0 to 1 of (arctan(2x)) ?

The integral from 0 to 1 of (arctan(2x)) is arctan(2)-1/4 ln(5)

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integral from 0 to 1 of (arctan(2x))

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

arctan (2) - \frac{1}{4} ln (5)

Decimal

0.70478 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

arctan (2) - \frac{1}{4} ln (5)

= arctan (2) - \frac{1}{4} ln (5)

f (x) = 7 x^{2} tan (x)

x y^{^{'}} + y = 7 x y^{2}

\int 6 x 3 x^{2} + 2 d x

f (x) = x + 7

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 2)

What is the integral from 0 to 1 of (arctan(2x)) ?
The integral from 0 to 1 of (arctan(2x)) is arctan(2)-1/4 ln(5)