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

The integral from 0 to 1 of (2y)/(1+x^2) is (piy)/2

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

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

\frac{π y}{2}

Solution steps

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

Take the constant out:

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

= 2 y \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)

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

Compute the boundaries:

\frac{π}{4}

= 2 y \frac{π}{4}

Simplify

= \frac{π y}{2}

\int_{1}^{\infty} x^{\frac{1}{2}} d x

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\int_{- 1}^{1} t^{2} cos (nπ t) d t

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

\int_{0}^{2} e^{x} d x

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