integral of (x-4)/((x+2)^2+16)

Lösung

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

\frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2)) + C

Schritte zur Lösung

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{u - 6}{u ^{2} + 16} d u

Multipliziere aus

\frac{u - 6}{u ^{2} + 16} : \frac{u}{u ^{2} + 16} - \frac{6}{u ^{2} + 16}

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{u}{u ^{2} + 16} d u - \int \frac{6}{u ^{2} + 16} d u

\int \frac{u}{u ^{2} + 16} d u = \frac{1}{2} ln u^{2} + 16

\int \frac{6}{u ^{2} + 16} d u = \frac{3}{2} arctan (\frac{u}{4})

= \frac{1}{2} ln u^{2} + 16 - \frac{3}{2} arctan (\frac{u}{4})

Setze in

u = x + 2

ein

= \frac{1}{2} ln (x + 2)^{2} + 16 - \frac{3}{2} arctan (\frac{x + 2}{4})

Vereinfache

\frac{1}{2} ln (x + 2)^{2} + 16 - \frac{3}{2} arctan (\frac{x + 2}{4}) : \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 4 x + 20 - \frac{3}{2} arctan (\frac{1}{4} (x + 2)) + C

x \to 0 lim (\frac{e ^{x^{3}} - 1}{x})

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