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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

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

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

Compute the boundaries:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Simplify

= \frac{π}{2}

\int \frac{1}{2 ( x ^{2} + 1 )} d x

\int \frac{1}{x ^{2}} x^{'} d x

(5 x - 3) y^{^{''}} + y^{^{'}} = 0

6

\frac{1}{x}, x = 2

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