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Beliebt Rechnen >

integral of (y+1)/(y^2+y+1)

Lösung

\int \frac{y + 1}{y ^{2} + y + 1} d y

Lösung

\frac{1}{2} ln 4 y^{2} + 4 y + 4 + \frac{1}{3} arctan (\frac{2 y + 1}{3}) + C

Schritte zur Lösung

\int \frac{y + 1}{y ^{2} + y + 1} d y

Vervollständige das Quadrat

y^{2} + y + 1 : (y + \frac{1}{2})^{2} + \frac{3}{4}

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= 2 \cdot \int \frac{2 u + 1}{4 u ^{2} + 3} d u

Multipliziere aus

\frac{2 u + 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{2} + 3}

Wende die Summenregel an:

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

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

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2 3} arctan (\frac{2}{3} u)

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

Setze in

u = y + \frac{1}{2}

ein

= 2 (\frac{1}{4} ln 4 (y + \frac{1}{2})^{2} + 3 + \frac{1}{2 3} arctan (\frac{2}{3} (y + \frac{1}{2})))

Vereinfache

2 (\frac{1}{4} ln 4 (y + \frac{1}{2})^{2} + 3 + \frac{1}{2 3} arctan (\frac{2}{3} (y + \frac{1}{2}))) : \frac{1}{2} ln 4 y^{2} + 4 y + 4 + \frac{1}{3} arctan (\frac{2 y + 1}{3})

= \frac{1}{2} ln 4 y^{2} + 4 y + 4 + \frac{1}{3} arctan (\frac{2 y + 1}{3})

Füge eine Konstante zur Lösung hinzu

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

Graph

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fläche e^{(x/4)},x=4,y=1

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fläche x^2,-1<= x<= 2

a re a x^{2}, - 1 \leq x \leq 2