What is the integral from-infinity to 0 of 1/(1+x^2) ?

The integral from-infinity to 0 of 1/(1+x^2) is pi/2

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integral from-infinity to 0 of 1/(1+x^2)

\int_{- \infty}^{0} \frac{1}{1 + x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{- \infty}^{0} \frac{1}{1 + x ^{2}} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{- \infty}^{0} \frac{1}{1 + x ^{2}} d x = 0 - (- \frac{π}{2})

= 0 - (- \frac{π}{2})

Simplify

= \frac{π}{2}

\frac{d}{d x} (ln (5 x + 1))

f (x) = \frac{19}{x ^{5}}

\frac{d}{d x} (- 2 lo g_{6} (x))

e^{9 x}

\frac{d y}{d x}, x y = 9 + x^{2} y

What is the integral from-infinity to 0 of 1/(1+x^2) ?
The integral from-infinity to 0 of 1/(1+x^2) is pi/2