integral from 0 to 1 of xln(x+3)

Lösung

\int_{0}^{1} x ln (x + 3) d x

- 8 ln (2) + \frac{5}{4} + \frac{9}{2} ln (3)

Dezimale

0.64857 \dots

Schritte zur Lösung

\int_{0}^{1} x ln (x + 3) d x

Wende U-Substitution an

= \int_{3}^{4} (u - 3) ln (u) d u

Wende die partielle Integration an

= [ln (u) (\frac{u ^{2}}{2} - 3 u) - \int \frac{1}{2} (u - 6) d u]_{3}^{4}

\int \frac{1}{2} (u - 6) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

= [ln (u) (\frac{u ^{2}}{2} - 3 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)]_{3}^{4}

Berechne die Grenzen:

- 8 ln (2) + \frac{5}{4} + \frac{9}{2} ln (3)

= - 8 ln (2) + \frac{5}{4} + \frac{9}{2} ln (3)

\int_{2}^{3} \frac{3}{x ^{2} - 5 x + 4} d x

\int_{- 1}^{0} x^{3} (4 x^{4} - 3)^{4} d x

\int_{4}^{7} \frac{5}{( x - 2 ) ( 2 x + 1 )} d x

\int_{0}^{3} (4 x^{3} + \frac{2}{x ^{3}}) d x

\int_{0}^{\frac{π}{3}} sin^{3} (θ) d θ