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

Lösung

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

- \frac{π}{4} + 1

Dezimale

0.21460 \dots

Schritte zur Lösung

\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

Wende die Summenregel an:

\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

\frac{d}{d x} (1 - \frac{2}{x})

y^{^{''}} + x y^{^{'}} - y = 0

\frac{d}{d x} (- \frac{cos ( x )}{x})

\int \frac{5 x ^{3}}{x ^{2} - 1} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = (1 + e^{t})^{- 1}