integral of 3x(4x-1)^5

Lösung

\int 3 x (4 x - 1)^{5} d x

\frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6}) + C

Schritte zur Lösung

\int 3 x (4 x - 1)^{5} d x

Entferne die Konstante:

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

= 3 \cdot \int x (4 x - 1)^{5} d x

Wende U-Substitution an

= 3 \cdot \int \frac{u ^{5} ( u + 1 )}{16} d u

Entferne die Konstante:

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

= 3 \cdot \frac{1}{16} \cdot \int u^{5} (u + 1) d u

Multipliziere aus

u^{5} (u + 1) : u^{6} + u^{5}

= 3 \cdot \frac{1}{16} \cdot \int u^{6} + u^{5} 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

= 3 \cdot \frac{1}{16} (\int u^{6} d u + \int u^{5} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{5} d u = \frac{u ^{6}}{6}

= 3 \cdot \frac{1}{16} (\frac{u ^{7}}{7} + \frac{u ^{6}}{6})

Setze in

u = 4 x - 1

ein

= 3 \cdot \frac{1}{16} (\frac{( 4 x - 1 ) ^{7}}{7} + \frac{( 4 x - 1 ) ^{6}}{6})

Vereinfache

3 \cdot \frac{1}{16} (\frac{( 4 x - 1 ) ^{7}}{7} + \frac{( 4 x - 1 ) ^{6}}{6}) : \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6})

= \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6}) + C

\int 7 sin^{3} (x) cos^{7} (x) d x

l a pl a ce t r an s f or m cos (3 t)

\frac{d}{d x} (7 ln (sec (x) + tan (x)))

t an g e n t y = 2 x^{3} - 6 x^{2} + 5 x + 1, (2, 3)

h^{^{'}} = - h