What is the integral from 0 to 1/2 of 2/(81+4x^2) ?

The integral from 0 to 1/2 of 2/(81+4x^2) is 1/9 arctan(1/9)

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integral from 0 to 1/2 of 2/(81+4x^2)

\int_{0}^{\frac{1}{2}} \frac{2}{81 + 4 x ^{2}} d x

\frac{1}{9} arctan (\frac{1}{9})

Decimal Notation

0.01229 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{2}{81 + 4 x ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{1}{2}} \frac{1}{81 + 4 x ^{2}} d x

Apply Integral Substitution

= 2 \cdot \int_{0}^{\frac{1}{9}} \frac{1}{18 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{18} \cdot \int_{0}^{\frac{1}{9}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 2 \cdot \frac{1}{18} [arctan (u)]_{0}^{\frac{1}{9}}

Simplify

2 \cdot \frac{1}{18} [arctan (u)]_{0}^{\frac{1}{9}} : \frac{1}{9} [arctan (u)]_{0}^{\frac{1}{9}}

= \frac{1}{9} [arctan (u)]_{0}^{\frac{1}{9}}

Compute the boundaries:

arctan (\frac{1}{9})

= \frac{1}{9} arctan (\frac{1}{9})

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What is the integral from 0 to 1/2 of 2/(81+4x^2) ?
The integral from 0 to 1/2 of 2/(81+4x^2) is 1/9 arctan(1/9)