integral of x/(sqrt(4x^2+4x+5))

Lösung

\int \frac{x}{4 x ^{2} + 4 x + 5} d x

\frac{1}{4} (4 x^{2} + 4 x + 5 - ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2}) + C

Schritte zur Lösung

\int \frac{x}{4 x ^{2} + 4 x + 5} d x

Vervollständige das Quadrat

4 x^{2} + 4 x + 5 : 4 (x + \frac{1}{2})^{2} + 4

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{2 u - 1}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{1}{u ^{2} + 1}

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

Wende die Summenregel an:

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

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

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

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

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

Setze in

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

ein

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

Vereinfache

\frac{1}{4} 2 (x + \frac{1}{2})^{2} + 1 - ln (x + \frac{1}{2})^{2} + 1 + x + \frac{1}{2} : \frac{1}{4} (4 x^{2} + 4 x + 5 - ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2})

= \frac{1}{4} (4 x^{2} + 4 x + 5 - ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (4 x^{2} + 4 x + 5 - ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2}) + C

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