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

The integral from 0 to 1 of ((x^2))/(x^2+1) is -pi/4+1

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

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

- \frac{π}{4} + 1

Decimal

0.21460 \dots

Solution steps

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

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

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

\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = \frac{π}{4}

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

= - \frac{π}{4} + 1

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What is the integral from 0 to 1 of ((x^2))/(x^2+1) ?
The integral from 0 to 1 of ((x^2))/(x^2+1) is -pi/4+1