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

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

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

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

\frac{π}{2} - 1

Decimal

0.57079 \dots

Solution steps

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

\frac{1 - x ^{2}}{1 + x ^{2}} = \frac{2}{1 + x ^{2}} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{2}{1 + x ^{2}} d x = \frac{π}{2}

\int_{0}^{1} 1 d x = 1

= \frac{π}{2} - 1

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