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

The integral from-infinity to 6 of 1/(x^2+1) is arctan(6)+pi/2

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

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

arctan (6) + \frac{π}{2}

Decimal Notation

2.97644 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{- \infty}^{6} \frac{1}{x ^{2} + 1} d x = arctan (6) - (- \frac{π}{2})

= arctan (6) - (- \frac{π}{2})

Simplify

= arctan (6) + \frac{π}{2}

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What is the integral from-infinity to 6 of 1/(x^2+1) ?
The integral from-infinity to 6 of 1/(x^2+1) is arctan(6)+pi/2