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

The integral from 0 to 1 of 1/(t^2+2) is (arctan(1/(sqrt(2))))/(sqrt(2))

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

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

\frac{arctan ( \frac{1}{2} )}{2}

Decimal Notation

0.43520 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{\frac{1}{2}} \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{0}^{\frac{1}{2}}

Compute the boundaries:

arctan (\frac{1}{2})

= \frac{1}{2} arctan (\frac{1}{2})

Simplify

= \frac{arctan ( \frac{1}{2} )}{2}

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