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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 4 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 4 \cdot \frac{π}{4}

Simplify

= π

\int_{0}^{\frac{π}{2}} cos (x) \frac{1}{2} d x

\int_{0}^{2} y 4 - y^{2} d y

\int_{0}^{4} (7 - x) x d x

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

\int_{1}^{2} \frac{y + 4 y ^{6}}{y ^{3}} d y

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