integral (4x^4)/((1+x^2))

Lösung

integral

\frac{4 x ^{4}}{( 1 + x ^{2} )}

4 (\frac{x ^{3}}{3} - x + arctan (x)) + C

Schritte zur Lösung

\int \frac{4 x ^{4}}{1 + x ^{2}} d x

Entferne die Konstante:

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

= 4 \cdot \int \frac{x ^{4}}{1 + x ^{2}} d x

Lange Division

\frac{x ^{4}}{1 + x ^{2}} : x^{2} - 1 + \frac{1}{x ^{2} + 1}

= 4 \cdot \int x^{2} - 1 + \frac{1}{x ^{2} + 1} 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

= 4 (\int x^{2} d x - \int 1 d x + \int \frac{1}{x ^{2} + 1} d x)

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

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

= 4 (\frac{x ^{3}}{3} - x + arctan (x))

Füge eine Konstante zur Lösung hinzu

= 4 (\frac{x ^{3}}{3} - x + arctan (x)) + C

in t e g r a l \frac{4 t}{3 t ^{2} - 7}

in t e g r a l u^{\frac{1}{2}}

in t e g r a l \frac{1}{( t ^{2} + t + 1 )}

in t e g r a l x^{2} e^{\frac{x ^{2}}{2}}

in t e g r a l - 2 x^{3} (e^{- x})^{2}