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

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

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

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

\frac{5 π}{4}

Decimal Notation

3.92699 \dots

Solution steps

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

Use the common integral:

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

= 5 [arctan (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 5 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{4}

(1 + \frac{y}{x ^{2}}) d x - \frac{1}{x} d y = 0

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

\frac{d}{d x} (- \frac{1}{4 sin ( x )})

\frac{1}{- 0.75 y - 0.00003} y^{^{'}} = 1

\int x^{3} x^{2} + 5 d x

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