What is the integral from 0 to 1/2 of 3/(1+4x^2) ?

The integral from 0 to 1/2 of 3/(1+4x^2) is (3pi)/8

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integral from 0 to 1/2 of 3/(1+4x^2)

\int_{0}^{\frac{1}{2}} \frac{3}{1 + 4 x ^{2}} d x

\frac{3 π}{8}

Solution steps

\int_{0}^{\frac{1}{2}} \frac{3}{1 + 4 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 x ^{2}} d x

Apply Integral Substitution

= 3 \cdot \int_{0}^{1} \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

3 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : \frac{3}{2} [arctan (u)]_{0}^{1}

= \frac{3}{2} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{3}{2} \cdot \frac{π}{4}

Simplify

= \frac{3 π}{8}

\frac{d}{d x} (\frac{( x - 5 x x )}{x})

t^{2} \frac{d y}{d t} + t y = 6

n = 0 \sum \infty \frac{n}{3} (\frac{2}{3})^{n}

\int_{1}^{5} \frac{x ^{2} - 1}{x} d x

What is the integral from 0 to 1/2 of 3/(1+4x^2) ?
The integral from 0 to 1/2 of 3/(1+4x^2) is (3pi)/8