integral of (x^4+2x-9)ln(x)

Lösung

\int (x^{4} + 2 x - 9) ln (x) d x

\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x) + C

Schritte zur Lösung

\int (x^{4} + 2 x - 9) ln (x) d x

Multipliziere aus

(x^{4} + 2 x - 9) ln (x) : x^{4} ln (x) + 2 x ln (x) - 9 ln (x)

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

Wende die Summenregel an:

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

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

\int x^{4} ln (x) d x = \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}

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

\int 9 ln (x) d x = 9 (x ln (x) - x)

= \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x) + C

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 3

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

y^{^{'}} + y^{2} sin (x) = 0

\frac{d y}{d x} = \frac{2 y}{2 x + 1}

t an g e n t 10 x^{2} + 2 x