What is the integral from 0 to 1/2 of 5/(1+4x^2) ?

The integral from 0 to 1/2 of 5/(1+4x^2) is (5pi)/8

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integral from 0 to 1/2 of 5/(1+4x^2)

\int_{0}^{\frac{1}{2}} \frac{5}{1 + 4 x ^{2}} d x

\frac{5 π}{8}

Decimal Notation

1.96349 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{5}{1 + 4 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 x ^{2}} d x

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

5 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : \frac{5}{2} [arctan (u)]_{0}^{1}

= \frac{5}{2} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{5}{2} \cdot \frac{π}{4}

Simplify

= \frac{5 π}{8}

e^{x} (3 x^{2} - x + 1)

\frac{\partial}{\partial y} (e^{- x y^{2}} + y^{3} x^{4})

x y

n = 1 \sum \infty n^{3} + n

(x^{5})^{^{'}}

What is the integral from 0 to 1/2 of 5/(1+4x^2) ?
The integral from 0 to 1/2 of 5/(1+4x^2) is (5pi)/8