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

Lösung

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

2 ln x^{2} + 9 + arctan (\frac{x}{3}) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

= 2 ln x^{2} + 9 + arctan (\frac{x}{3})

Füge eine Konstante zur Lösung hinzu

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

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 x ^{5}})

\int_{2}^{4} (\frac{5}{x} - \frac{10}{x ^{2}}) d x

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