integral of arctan(4t)

Lösung

\int arctan (4 t) d t

t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1 + C

Schritte zur Lösung

\int arctan (4 t) d t

Wende die partielle Integration an

= t arctan (4 t) - \int \frac{4 t}{16 t ^{2} + 1} d t

\int \frac{4 t}{16 t ^{2} + 1} d t = \frac{1}{8} ln 16 t^{2} + 1

= t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1

Füge eine Konstante zur Lösung hinzu

= t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1 + C

n = 3 \sum \infty \frac{1}{ln ( n )}

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{5})

\int_{1}^{3} (4 x^{3} - 3 x^{2}) d x

in v erse l a pl a ce \frac{1}{s ^{2} + s}

n = 1 \sum \infty - \frac{1}{n ^{2}}