integral of ((5x+3))/((x^2+4x+7))

Lösung

\int \frac{( 5 x + 3 )}{( x ^{2} + 4 x + 7 )} d x

\frac{5}{2} ln \frac{( x + 2 ) ^{2}}{3} + 1 - \frac{7}{3} arctan (\frac{1}{3} (x + 2)) + C

Schritte zur Lösung

\int \frac{5 x + 3}{x ^{2} + 4 x + 7} d x

Vervollständige das Quadrat

x^{2} + 4 x + 7 : (x + 2)^{2} + 3

= \int \frac{5 x + 3}{( x + 2 ) ^{2} + 3} d x

Wende U-Substitution an

= \int \frac{5 u - 7}{u ^{2} + 3} d u

Wende integrale Substitution an

= \int \frac{5 3 v - 7}{3 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{5 3 v - 7}{v ^{2} + 1} d v

Multipliziere aus

\frac{5 3 v - 7}{v ^{2} + 1} : \frac{5 3 v}{v ^{2} + 1} - \frac{7}{v ^{2} + 1}

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}{3} (\int \frac{5 3 v}{v ^{2} + 1} d v - \int \frac{7}{v ^{2} + 1} d v)

\int \frac{5 3 v}{v ^{2} + 1} d v = \frac{5 3}{2} ln v^{2} + 1

\int \frac{7}{v ^{2} + 1} d v = 7 arctan (v)

= \frac{1}{3} (\frac{5 3}{2} ln v^{2} + 1 - 7 arctan (v))

Ersetze zurück

= \frac{1}{3} (\frac{5 3}{2} ln (\frac{x + 2}{3})^{2} + 1 - 7 arctan (\frac{x + 2}{3}))

Vereinfache

\frac{1}{3} (\frac{5 3}{2} ln (\frac{x + 2}{3})^{2} + 1 - 7 arctan (\frac{x + 2}{3})) : \frac{5}{2} ln \frac{( x + 2 ) ^{2}}{3} + 1 - \frac{7}{3} arctan (\frac{1}{3} (x + 2))

= \frac{5}{2} ln \frac{( x + 2 ) ^{2}}{3} + 1 - \frac{7}{3} arctan (\frac{1}{3} (x + 2))

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} ln \frac{( x + 2 ) ^{2}}{3} + 1 - \frac{7}{3} arctan (\frac{1}{3} (x + 2)) + C

\frac{d}{d x} ((5 x - 1)^{2})

k = 1 \sum \infty \frac{k}{k ^{2} + 1}

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