integral of ln(x+1)x

Lösung

\int ln (x + 1) x 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 ln (x + 1) x 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

\frac{\partial}{\partial v} (\frac{u}{u + v})

y^{^{'}} = e^{6 x} - 4 x

t an g e n t f (x) = x^{2} - 3 x + 1, at x = 1

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{3} + 4 y)

d er i v a t i v e \frac{- 200 ln ( \frac{f}{40} )}{( f + 20 )}