integral from 0 to 1 of (2x)/(x^2+4x+29)

Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 4 x + 29} d x

\frac{5 ( ln ( \frac{34}{25} ) - ln ( \frac{29}{25} ) ) - 4 ( arctan ( \frac{3}{5} ) - arctan ( \frac{2}{5} ) )}{5}

Dezimale

0.03113 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{2 x}{x ^{2} + 4 x + 29} d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} \frac{x}{x ^{2} + 4 x + 29} d x

Vervollständige das Quadrat

x^{2} + 4 x + 29 : (x + 2)^{2} + 25

= 2 \cdot \int_{0}^{1} \frac{x}{( x + 2 ) ^{2} + 25} d x

Wende U-Substitution an

= 2 \cdot \int_{2}^{3} \frac{u - 2}{u ^{2} + 25} d u

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

\int_{\frac{2}{5}}^{\frac{3}{5}} \frac{5 v}{v ^{2} + 1} d v = \frac{5}{2} (ln (\frac{34}{25}) - ln (\frac{29}{25}))

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

= 2 \cdot \frac{1}{5} (\frac{5}{2} (ln (\frac{34}{25}) - ln (\frac{29}{25})) - 2 (arctan (\frac{3}{5}) - arctan (\frac{2}{5})))

Vereinfache

2 \cdot \frac{1}{5} (\frac{5}{2} (ln (\frac{34}{25}) - ln (\frac{29}{25})) - 2 (arctan (\frac{3}{5}) - arctan (\frac{2}{5}))) : \frac{5 ( ln ( \frac{34}{25} ) - ln ( \frac{29}{25} ) ) - 4 ( arctan ( \frac{3}{5} ) - arctan ( \frac{2}{5} ) )}{5}

= \frac{5 ( ln ( \frac{34}{25} ) - ln ( \frac{29}{25} ) ) - 4 ( arctan ( \frac{3}{5} ) - arctan ( \frac{2}{5} ) )}{5}

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