What is the integral from 1 to sqrt(3 of) 2/(x^2+1) ?

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

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integral from 1 to sqrt(3 of) 2/(x^2+1)

\int_{1}^{3} \frac{2}{x ^{2} + 1} d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{1}^{3} \frac{2}{x ^{2} + 1} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{3} \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 2 [arctan (x)]_{1}^{3}

Compute the boundaries:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Simplify

= \frac{π}{6}

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\int_{0}^{4} \frac{1}{x ^{3}} d x

\int_{0}^{\frac{π}{5}} x cos (5 x) d x

\int_{0}^{7} tan (x) d x

\int_{- 1}^{2} (x - 1)^{2} d x

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