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

The integral from 1 to 2 of 1/(1+2x^2) is 1/(sqrt(2))(arctan(sqrt(2)*2)-arctan(sqrt(2)))

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

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

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

Decimal Notation

0.19490 \dots

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

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

Compute the boundaries:

arctan (2 \cdot 2) - arctan (2)

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

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What is the integral from 1 to 2 of 1/(1+2x^2) ?
The integral from 1 to 2 of 1/(1+2x^2) is 1/(sqrt(2))(arctan(sqrt(2)*2)-arctan(sqrt(2)))