integral of (x+1)/(\sqrt[4]{x)}(x-1)

Lösung

\int \frac{x + 1}{4 x} (x - 1) d x

\frac{4}{11} x^{\frac{11}{4}} - \frac{4}{3} x^{\frac{3}{4}} + C

Schritte zur Lösung

\int \frac{x + 1}{4 x} (x - 1) d x

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int \frac{( x + 1 ) ( x - 1 )}{4 x} d x

Multipliziere aus

\frac{( x + 1 ) ( x - 1 )}{4 x} : x^{\frac{7}{4}} - \frac{1}{4 x}

= \int x^{\frac{7}{4}} - \frac{1}{4 x} 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 x^{\frac{7}{4}} d x - \int \frac{1}{4 x} d x

\int x^{\frac{7}{4}} d x = \frac{4}{11} x^{\frac{11}{4}}

\int \frac{1}{4 x} d x = \frac{4}{3} x^{\frac{3}{4}}

= \frac{4}{11} x^{\frac{11}{4}} - \frac{4}{3} x^{\frac{3}{4}}

Füge eine Konstante zur Lösung hinzu

= \frac{4}{11} x^{\frac{11}{4}} - \frac{4}{3} x^{\frac{3}{4}} + C

\int_{1}^{3} 13 r^{3} ln (r) d r

n = 0 \sum \infty 5 n

\int \frac{( 7 ( 1 - tan ^{2} ( x ) ) )}{sec ^{2} ( x )} d x

\frac{\partial}{\partial x} (ln (4 x^{2} + 7 y^{2} + 4))

\int x^{2} I n (x) d x