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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= 4 \cdot \frac{π}{4}

Simplify

= π

ln (1 + x^{2})

x^{2} + 7, 2 x^{2} + 2 x - 1

\int \frac{1}{9 x ^{2} - 16} d x

\int \frac{4 x ^{2}}{x ^{2} + 6} d x

y^{^{'}} - y = 3

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