integral of (5x^3+7x+2)/(x^4+9x^2)

Lösung

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

\frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C

Schritte zur Lösung

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

Multipliziere aus

\frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} : \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} + \frac{7 x}{x ^{4} + 9 x ^{2}} + \frac{2}{x ^{4} + 9 x ^{2}}

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 x ^{3}}{x ^{4} + 9 x ^{2}} d x + \int \frac{7 x}{x ^{4} + 9 x ^{2}} d x + \int \frac{2}{x ^{4} + 9 x ^{2}} d x

\int \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} d x = \frac{5}{2} ln x^{2} + 9

\int \frac{7 x}{x ^{4} + 9 x ^{2}} d x = 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9)

\int \frac{2}{x ^{4} + 9 x ^{2}} d x = 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

= \frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

Vereinfache

\frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})) : \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

Füge eine Konstante zur Lösung hinzu

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C

in v erse l a pl a ce \frac{2}{( s + 1 ) ( s + 2 )}

\frac{\partial}{\partial u} (sin^{2} (u))

t an g e n t 18 cos (x) + 9 sin (2 x)

\int_{0}^{4 π} \frac{1}{2}

n = 1 \sum \infty \frac{n ^{2}}{6 ^{n}}