integral of x(x^2+1)^{100}

Lösung

\int x (x^{2} + 1)^{100} d x

\frac{1}{202} (x^{2} + 1)^{101} + C

Schritte zur Lösung

\int x (x^{2} + 1)^{100} d x

Wende U-Substitution an

= \int \frac{u ^{100}}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int u^{100} d u

Wende die Potenzregel an

= \frac{1}{2} \cdot \frac{u ^{101}}{101}

Setze in

u = x^{2} + 1

ein

= \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101}

Vereinfache

\frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101} : \frac{1}{202} (x^{2} + 1)^{101}

= \frac{1}{202} (x^{2} + 1)^{101}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{202} (x^{2} + 1)^{101} + C

l o n g d i v i s i o n \frac{x ^{2} - 10 x + 65}{x - 8}

n = 0 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}

x \to 5 lim (\frac{sin ( 5 π )}{3})

\frac{\partial}{\partial x} (x^{2} + 8 y^{2} + 9 z^{2} - 7)

\frac{d}{d x} (f (x) (5 x^{2} + 3))