integral of t+2+3/(t^2+1)

Lösung

\int t + 2 + \frac{3}{t ^{2} + 1} d t

\frac{t ^{2}}{2} + 2 t + 3 arctan (t) + C

Schritte zur Lösung

\int t + 2 + \frac{3}{t ^{2} + 1} d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t d t + \int 2 d t + \int \frac{3}{t ^{2} + 1} d t

\int t d t = \frac{t ^{2}}{2}

\int 2 d t = 2 t

\int \frac{3}{t ^{2} + 1} d t = 3 arctan (t)

= \frac{t ^{2}}{2} + 2 t + 3 arctan (t)

Füge eine Konstante zur Lösung hinzu

= \frac{t ^{2}}{2} + 2 t + 3 arctan (t) + C

\frac{d}{d x} (\frac{- 8 x ^{7} + 7 x ^{3} - 2}{- 4 x ^{3}})

\int (a x + b + x^{2})^{2} d x

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

\frac{d}{d x} ((- x^{2} + 4 x + 5) (4 x^{4} - 3))

\frac{d y}{d x} - \frac{2 y}{x} = x^{2} - 1