integral of (10)/((r-1)^2(r^2+9))

Lösung

\int \frac{10}{( r - 1 ) ^{2} ( r ^{2} + 9 )} d r

- \frac{1}{5} ln ∣ r - 1 ∣ - \frac{1}{r - 1} + \frac{1}{10} ln r^{2} + 9 - \frac{4}{15} arctan (\frac{r}{3}) + C

Schritte zur Lösung

\int \frac{10}{( r - 1 ) ^{2} ( r ^{2} + 9 )} d r

Entferne die Konstante:

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

= 10 \cdot \int \frac{1}{( r - 1 ) ^{2} ( r ^{2} + 9 )} d r

Ermittle den Partialbruch von

\frac{1}{( r - 1 ) ^{2} ( r ^{2} + 9 )} : - \frac{1}{50 ( r - 1 )} + \frac{1}{10 ( r - 1 ) ^{2}} + \frac{r - 4}{50 ( r ^{2} + 9 )}

= 10 \cdot \int - \frac{1}{50 ( r - 1 )} + \frac{1}{10 ( r - 1 ) ^{2}} + \frac{r - 4}{50 ( r ^{2} + 9 )} d r

Wende die Summenregel an:

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

= 10 (- \int \frac{1}{50 ( r - 1 )} d r + \int \frac{1}{10 ( r - 1 ) ^{2}} d r + \int \frac{r - 4}{50 ( r ^{2} + 9 )} d r)

\int \frac{1}{50 ( r - 1 )} d r = \frac{1}{50} ln ∣ r - 1 ∣

\int \frac{1}{10 ( r - 1 ) ^{2}} d r = - \frac{1}{10 ( r - 1 )}

\int \frac{r - 4}{50 ( r ^{2} + 9 )} d r = \frac{1}{50} (\frac{1}{2} ln r^{2} + 9 - \frac{4}{3} arctan (\frac{r}{3}))

= 10 (- \frac{1}{50} ln ∣ r - 1 ∣ - \frac{1}{10 ( r - 1 )} + \frac{1}{50} (\frac{1}{2} ln r^{2} + 9 - \frac{4}{3} arctan (\frac{r}{3})))

Vereinfache

10 (- \frac{1}{50} ln ∣ r - 1 ∣ - \frac{1}{10 ( r - 1 )} + \frac{1}{50} (\frac{1}{2} ln r^{2} + 9 - \frac{4}{3} arctan (\frac{r}{3}))) : - \frac{1}{5} ln ∣ r - 1 ∣ - \frac{1}{r - 1} + \frac{1}{10} ln r^{2} + 9 - \frac{4}{15} arctan (\frac{r}{3})

= - \frac{1}{5} ln ∣ r - 1 ∣ - \frac{1}{r - 1} + \frac{1}{10} ln r^{2} + 9 - \frac{4}{15} arctan (\frac{r}{3})

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{5} ln ∣ r - 1 ∣ - \frac{1}{r - 1} + \frac{1}{10} ln r^{2} + 9 - \frac{4}{15} arctan (\frac{r}{3}) + C

\int (x^{2} - 9 x - 1) d x

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\int (\frac{3}{x ^{3}} - \frac{1}{3 x}) d x

\int \frac{t ^{2}}{1 - t ^{2}} d t

\int \frac{sin ( x )}{( sin ( x ) + cos ( x ) )} d x