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

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

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

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

- \frac{π}{2} + 2

Decimal

0.42920 \dots

Solution steps

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

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

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

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

\int_{- 1}^{1} 1 d x = 2

= - \frac{π}{2} + 2

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