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

Lösung

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

\frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{5 x ^{3} - 3 x ^{2} + 4 x - 3}{x ^{4} + x ^{2}} : \frac{5 x ^{3}}{x ^{4} + x ^{2}} - \frac{3 x ^{2}}{x ^{4} + x ^{2}} + \frac{4 x}{x ^{4} + x ^{2}} - \frac{3}{x ^{4} + 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} + x ^{2}} d x - \int \frac{3 x ^{2}}{x ^{4} + x ^{2}} d x + \int \frac{4 x}{x ^{4} + x ^{2}} d x - \int \frac{3}{x ^{4} + x ^{2}} d x

\int \frac{5 x ^{3}}{x ^{4} + x ^{2}} d x = \frac{5}{2} ln x^{2} + 1

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

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

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

= \frac{5}{2} ln x^{2} + 1 - 3 arctan (x) + 4 (- \frac{1}{2} ln x^{2} + 1 + ln (x)) - 3 (- \frac{1}{x} - arctan (x))

Vereinfache

\frac{5}{2} ln x^{2} + 1 - 3 arctan (x) + 4 (- \frac{1}{2} ln x^{2} + 1 + ln (x)) - 3 (- \frac{1}{x} - arctan (x)) : \frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x}

= \frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x} + C

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

\int \frac{1}{2 - 3 x - x ^{2}} d x

\int (5 x + 6)^{4} d x

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

\int \frac{cos ( x )}{5 sin ^{2} ( x ) + 6 sin ( x )} d x