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

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

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

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

- \frac{π}{4} + 1

Decimal Notation

0.21460 \dots

Solution steps

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

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

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

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

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

= - \frac{π}{4} + 1

\frac{d}{d x} (\frac{5 - 3 x}{5 + 3 x})

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

\int_{9}^{10} x x - 9 d x

\int e^{x^{5}} d x

y = (2 x + 5)^{5} (4 x + 5)^{- 4}

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