What is the integral from 0 to 1 of (6/(x^2+1)) ?

The integral from 0 to 1 of (6/(x^2+1)) is (3pi)/2

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integral from 0 to 1 of (6/(x^2+1))

\int_{0}^{1} (\frac{6}{x ^{2} + 1}) d x

\frac{3 π}{2}

Decimal

4.71238 \dots

Solution steps

\int_{0}^{1} \frac{6}{x ^{2} + 1} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 6 [arctan (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 6 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{2}

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

\int_{0}^{π} (sec (x))^{2} d x

What is the integral from 0 to 1 of (6/(x^2+1)) ?
The integral from 0 to 1 of (6/(x^2+1)) is (3pi)/2