integral from 0 to x of ln(1+2t)

Lösung

\int_{0}^{x} ln (1 + 2 t) d t

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

Schritte zur Lösung

\int_{0}^{x} ln (1 + 2 t) d t

Wende U-Substitution an

= \int_{1}^{1 + 2 x} ln (u) \frac{1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{1 + 2 x} ln (u) d u

Wende die partielle Integration an

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

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{1}^{1 + 2 x}

Berechne die Grenzen:

ln (2 x + 1) + 2 x ln (2 x + 1) - 2 x

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

Vereinfache

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

\int_{2}^{4} 3^{3 x} d x

\int_{- 1}^{1} \frac{x ^{2} - 1}{x ^{2} - 4} d x

\int_{0}^{1} \frac{x ^{2}}{4} d x

\int_{1}^{5} \frac{2}{x ^{2}} + x d x

\int_{1}^{3} \frac{x}{( x ^{2} + 9 ) ^{3}} d x