integral of tln(sqrt(t+1))

Lösung

\int t ln (t + 1) d t

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

Schritte zur Lösung

\int t ln (t + 1) d t

Vereinfache

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

= \int \frac{1}{2} t ln (t + 1) d t

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Wende die partielle Integration an

= \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

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

Setze in

u = t + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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