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

The integral from-infinity to 1 of 1/(x^2+1) is (3pi)/4

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

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

\frac{3 π}{4}

Decimal Notation

2.35619 \dots

Solution steps

\int_{- \infty}^{1} \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}^{1} \frac{1}{x ^{2} + 1} d x = \frac{π}{4} - (- \frac{π}{2})

= \frac{π}{4} - (- \frac{π}{2})

Simplify

= \frac{3 π}{4}

\int 2 x (x^{2} - 1)^{97} d x

\frac{d}{d x} (tan^{x} (x))

sin (\frac{π}{x})

12 y y^{^{'}} - 7 e^{x} = 0

\frac{\partial}{\partial x} ((sin (x))^{2})

What is the integral from-infinity to 1 of 1/(x^2+1) ?
The integral from-infinity to 1 of 1/(x^2+1) is (3pi)/4