integral of 1/((x+6)(36x^2+6x+1))

Lösung

\int \frac{1}{( x + 6 ) ( 36 x ^{2} + 6 x + 1 )} d x

16 (\frac{1}{20176} ln ∣ 12 x + 72 ∣ + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + \frac{1}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + \frac{1}{12})))) + C

Schritte zur Lösung

\int \frac{1}{( x + 6 ) ( 36 x ^{2} + 6 x + 1 )} d x

Wende U-Substitution an

= \int \frac{1}{u ( 36 u ^{2} - 426 u + 1261 )} d u

Vervollständige das Quadrat

36 u^{2} - 426 u + 1261 : 36 (u - \frac{71}{12})^{2} + \frac{3}{4}

= \int \frac{1}{u ( 36 ( u - \frac{71}{12} ) ^{2} + \frac{3}{4} )} d u

Wende U-Substitution an

= \int \frac{16}{( 12 v + 71 ) ( 48 v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 16 \cdot \int \frac{1}{( 12 v + 71 ) ( 48 v ^{2} + 1 )} d v

Ermittle den Partialbruch von

\frac{1}{( 12 v + 71 ) ( 48 v ^{2} + 1 )} : \frac{3}{5044 ( 12 v + 71 )} + \frac{- 12 v + 71}{5044 ( 48 v ^{2} + 1 )}

= 16 \cdot \int \frac{3}{5044 ( 12 v + 71 )} + \frac{- 12 v + 71}{5044 ( 48 v ^{2} + 1 )} d v

Wende die Summenregel an:

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

= 16 (\int \frac{3}{5044 ( 12 v + 71 )} d v + \int \frac{- 12 v + 71}{5044 ( 48 v ^{2} + 1 )} d v)

\int \frac{3}{5044 ( 12 v + 71 )} d v = \frac{1}{20176} ln ∣ 12 v + 71 ∣

\int \frac{- 12 v + 71}{5044 ( 48 v ^{2} + 1 )} d v = \frac{1}{5044} (- \frac{1}{8} ln 48 v^{2} + 1 + \frac{71}{4 3} arctan (43 v))

= 16 (\frac{1}{20176} ln ∣ 12 v + 71 ∣ + \frac{1}{5044} (- \frac{1}{8} ln 48 v^{2} + 1 + \frac{71}{4 3} arctan (43 v)))

Ersetze zurück

= 16 (\frac{1}{20176} ln 12 (x + 6 - \frac{71}{12}) + 71 + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + 6 - \frac{71}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + 6 - \frac{71}{12}))))

Vereinfache

16 (\frac{1}{20176} ln 12 (x + 6 - \frac{71}{12}) + 71 + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + 6 - \frac{71}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + 6 - \frac{71}{12})))) : 16 (\frac{1}{20176} ln ∣ 12 x + 72 ∣ + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + \frac{1}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + \frac{1}{12}))))

= 16 (\frac{1}{20176} ln ∣ 12 x + 72 ∣ + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + \frac{1}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + \frac{1}{12}))))

Füge eine Konstante zur Lösung hinzu

= 16 (\frac{1}{20176} ln ∣ 12 x + 72 ∣ + \frac{1}{5044} (- \frac{1}{8} ln 48 (x + \frac{1}{12})^{2} + 1 + \frac{71}{4 3} arctan (43 (x + \frac{1}{12})))) + C