integral of sqrt(x+x^4)

Lösung

\int x + x^{4} d x

\frac{1}{3} (x^{\frac{3}{2}} 1 + x^{3} + ln x^{\frac{3}{2}} + 1 + x^{3}) + C

Schritte zur Lösung

\int x + x^{4} d x

Faktorisiere

x + x^{4} : x (1 + x^{3})

= \int x (1 + x^{3}) d x

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (1 + x^{3}) = x (1 + x^{3})

= \int x 1 + x^{3} d x

Wende U-Substitution an

= \int \frac{2 u ^{2}}{3 u ^{2} - 1} d u

Entferne die Konstante:

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

= \frac{2}{3} \cdot \int \frac{u ^{2}}{u ^{2} - 1} d u

Trigonometrische Substitution anwenden

= \frac{2}{3} \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{2}{3} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{2}{3} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= \frac{2}{3} (\frac{sec ^{2} ( arcsec ( 1 + x ^{3} ) ) sin ( arcsec ( 1 + x ^{3} ) )}{2} + \frac{1}{2} ln tan (arcsec (1 + x^{3})) + sec (arcsec (1 + x^{3})))

Vereinfache

\frac{2}{3} (\frac{sec ^{2} ( arcsec ( 1 + x ^{3} ) ) sin ( arcsec ( 1 + x ^{3} ) )}{2} + \frac{1}{2} ln tan (arcsec (1 + x^{3})) + sec (arcsec (1 + x^{3}))) : \frac{1}{3} (x^{\frac{3}{2}} 1 + x^{3} + ln x^{\frac{3}{2}} + 1 + x^{3})

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (x^{\frac{3}{2}} 1 + x^{3} + ln x^{\frac{3}{2}} + 1 + x^{3}) + C

\int x e^{(x^{2} - 1)} d x

\int \frac{4}{( 1 + 2 x ^{2} ) ^{3}} d x

\int \frac{4 x ^{2} + x + 17}{( x + 1 ) ( x ^{2} + 4 )} d x

\int (x - 1) (2 - x) d x

\int \frac{( x ^{2} - 2 )}{x ^{3} - 6 x + 2} d x