integral of (x+1)sqrt(6-x)

Lösung

\int (x + 1) 6 - x d x

\frac{2}{5} (6 - x)^{\frac{5}{2}} - \frac{14}{3} (6 - x)^{\frac{3}{2}} + C

Schritte zur Lösung

\int (x + 1) 6 - x d x

Wende U-Substitution an

= \int - u (- u + 7) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u (- u + 7) d u

Multipliziere aus

u (- u + 7) : - u^{\frac{3}{2}} + 7 u

= - \int - u^{\frac{3}{2}} + 7 u d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{\frac{3}{2}} d u + \int 7 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 7 u d u = \frac{14}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{14}{3} u^{\frac{3}{2}})

Setze in

u = 6 - x

ein

= - (- \frac{2}{5} (6 - x)^{\frac{5}{2}} + \frac{14}{3} (6 - x)^{\frac{3}{2}})

Vereinfache

= \frac{2}{5} (6 - x)^{\frac{5}{2}} - \frac{14}{3} (6 - x)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{2}{5} (6 - x)^{\frac{5}{2}} - \frac{14}{3} (6 - x)^{\frac{3}{2}} + C

x \to - 3 - lim (x^{3} + 27)

f (x) = \frac{1}{3 x ^{2}}

t a y l or lo g_{10} (x + 1)

\frac{d}{d l} (\frac{d l ^{2} - d - a l ^{3} - b l ^{2}}{l ^{2}})

\int \frac{sin ( 2 x )}{( 1 - cos ^{2} ( x ) ) ^{3}} d x