integral from 0 to 1 of (x^3)/(1+x^3)

Lösung

\int_{0}^{1} \frac{x ^{3}}{1 + x ^{3}} d x

- \frac{1}{3} ln (2) - \frac{π}{3 3} + 1

Dezimale

0.16435 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{x ^{3}}{1 + x ^{3}} d x

\frac{x ^{3}}{1 + x ^{3}} = - \frac{1}{1 + x ^{3}} + 1

= \int_{0}^{1} - \frac{1}{1 + x ^{3}} + 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}{1 + x ^{3}} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} \frac{1}{1 + x ^{3}} d x = \frac{1}{3} ln (2) + \frac{π}{3 3}

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

= - (\frac{1}{3} ln (2) + \frac{π}{3 3}) + 1

Vereinfache

= - \frac{1}{3} ln (2) - \frac{π}{3 3} + 1

\frac{d}{d t} (- 2 cos (t))

y^{^{''}} = - a y

\frac{d}{d x} (- 2 cos (- 3 x + 7))

\int e^{2 x + 7} d x

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