integral of x*ln(x-1)

Lösung

\int x \cdot ln (x - 1) d x

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

Schritte zur Lösung

\int x ln (x - 1) d x

Wende U-Substitution an

= \int (u + 1) ln (u) d u

Wende die partielle Integration an

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

\int \frac{1}{2} (u + 2) d u = \frac{u ^{2}}{4} + u

= ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)

Setze in

u = x - 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

2 x \frac{d y}{d x} - y = 3 x^{2}

x \to 2 lim (\frac{1}{2 x})

\int 3 x^{2} e^{9 x^{3}} d x

d er i v a t i v e f (x) = \frac{x - 1}{x + 3}

\int_{0}^{\infty} e^{(- 2 x)} d x