integral of sqrt((2x+1)^5)

Lösung

\int (2 x + 1)^{5} d x

(2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}}) + C

Schritte zur Lösung

\int (2 x + 1)^{5} d x

(2 x + 1)^{5} = (2 x + 1)^{\frac{5}{2}},

angenommen

(2 x + 1) \geq 0

= \int (2 x + 1)^{\frac{5}{2}} d x

Wende die partielle Integration an

= (2 x + 1)^{\frac{5}{2}} x - \int 5 (2 x + 1)^{\frac{3}{2}} x d x

\int 5 (2 x + 1)^{\frac{3}{2}} x d x = 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}})

= (2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}})

Füge eine Konstante zur Lösung hinzu

= (2 x + 1)^{\frac{5}{2}} x - 5 (\frac{2}{7} x^{2} (2 x + 1)^{\frac{3}{2}} + \frac{3}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{1}{35} (2 x + 1)^{\frac{3}{2}}) + C

in v erse l a pl a ce \frac{- 2}{( s ^{2} + 1 ) ^{2}}

\int_{0}^{π} sin^{4} (3 t) d t

\int 40 x^{3} + 20 d x

\int 20 - (20)^{2} - x^{2} d x

\int (4 x^{3} + 3^{x}) d x