integral of (x^2+5x+4)/(x^2-2x+1)

Lösung

\int \frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} d x

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

Schritte zur Lösung

\int \frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} d x

Multipliziere aus

\frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} : \frac{x ^{2}}{x ^{2} - 2 x + 1} + \frac{5 x}{x ^{2} - 2 x + 1} + \frac{4}{x ^{2} - 2 x + 1}

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{x ^{2}}{x ^{2} - 2 x + 1} d x + \int \frac{5 x}{x ^{2} - 2 x + 1} d x + \int \frac{4}{x ^{2} - 2 x + 1} d x

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

\int \frac{5 x}{x ^{2} - 2 x + 1} d x = 5 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

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

Vereinfache

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + 5 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{4}{x - 1} : \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 7 ln ∣ x - 1 ∣ - \frac{13}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

\frac{d}{d x} (tan (sec (cos (x))))

s l o p e y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

d er i v a t i v e (\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 5 y^{3})

\frac{d y}{d x} = (e^{x - y})

\int_{2}^{x^{3}} t^{4} d t