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Beliebt Rechnen >

integral of (5x+1)/(sqrt(4x^2+4x+2))

Lösung

\int \frac{5 x + 1}{4 x ^{2} + 4 x + 2} d x

Lösung

\frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2 + C

Schritte zur Lösung

\int \frac{5 x + 1}{4 x ^{2} + 4 x + 2} d x

Vervollständige das Quadrat

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

= \int \frac{5 x + 1}{4 ( x + \frac{1}{2} ) ^{2} + 1} d x

Wende U-Substitution an

= \int \frac{5 ( u - \frac{1}{2} ) + 1}{4 u ^{2} + 1} d u

Multipliziere aus

\frac{5 ( u - \frac{1}{2} ) + 1}{4 u ^{2} + 1} : \frac{5 u}{4 u ^{2} + 1} - \frac{3}{2 4 u ^{2} + 1}

= \int \frac{5 u}{4 u ^{2} + 1} - \frac{3}{2 4 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

= \int \frac{5 u}{4 u ^{2} + 1} d u - \int \frac{3}{2 4 u ^{2} + 1} d u

\int \frac{5 u}{4 u ^{2} + 1} d u = \frac{5}{4} 4 u^{2} + 1

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

= \frac{5}{4} 4 u^{2} + 1 - \frac{3}{4} ln 2 u + 1 + 4 u^{2}

Setze in

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

ein

= \frac{5}{4} 4 (x + \frac{1}{2})^{2} + 1 - \frac{3}{4} ln 2 (x + \frac{1}{2}) + 1 + 4 (x + \frac{1}{2})^{2}

Vereinfache

\frac{5}{4} 4 (x + \frac{1}{2})^{2} + 1 - \frac{3}{4} ln 2 (x + \frac{1}{2}) + 1 + 4 (x + \frac{1}{2})^{2} : \frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2

= \frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2

Füge eine Konstante zur Lösung hinzu

= \frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2 + C

Graph

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