integral of 5xln(sqrt(1+x))

Lösung

\int 5 x ln (1 + x) d x

\frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3) + C

Schritte zur Lösung

\int 5 x ln (1 + x) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int x ln (1 + x) d x

Vereinfache

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

= 5 \cdot \int \frac{1}{2} x ln (1 + x) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \frac{1}{2} \cdot \int x ln (1 + x) d x

Wende U-Substitution an

= 5 \cdot \frac{1}{2} \cdot \int (u - 1) ln (u) d u

Wende die partielle Integration an

= 5 \cdot \frac{1}{2} (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

= 5 \cdot \frac{1}{2} (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Setze in

u = 1 + x

ein

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

Vereinfache

5 \cdot \frac{1}{2} (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x))) : \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3)

= \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3)

Füge eine Konstante zur Lösung hinzu

= \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3) + C

a re a x^{2}, x

\frac{d}{d x} (8 sin (8 x))

\frac{d}{d x} (x^{\frac{1}{1}})

x \to + 0 + lim (\frac{x ^{2}}{4} - \frac{2}{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = 1 + x