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

Lösung

integral

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

\frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{x ^{2}}{x + 4} d x = \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣

\int \frac{2}{x + 4} d x = 2 ln ∣ x + 4 ∣

= \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ - 2 ln ∣ x + 4 ∣

Vereinfache

\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ - 2 ln ∣ x + 4 ∣ : \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣

= \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣ + C

in t e g r a l (\frac{1}{x}) sin (\frac{1}{x})

in t e g r a l \frac{- 2}{y}

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in t e g r a l \frac{( x ^{24} )}{( x ^{25} )}

in t e g r a l \frac{( 2 v + 5 )}{( 7 v + 22 )}