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

The integral from sqrt(3) to 1 of 1/(x^2+1) is -pi/(12)

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

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

- \frac{π}{12}

Decimal

- 0.26179 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{12}

= - \frac{π}{12}

\int_{0}^{5} x^{2} - x d x

\int_{0}^{3} \frac{x e ^{x^{2}}}{3} d x

\int_{- 1}^{0} e^{- x} - e^{x} d x

\int_{- 9}^{0} (1 + 81 - x^{2}) d x

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

What is the integral from sqrt(3) to 1 of 1/(x^2+1) ?
The integral from sqrt(3) to 1 of 1/(x^2+1) is -pi/(12)