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

Lösung

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

- arctan (t) + t + C

Schritte zur Lösung

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

\frac{t ^{2}}{1 + t ^{2}} = - \frac{1}{1 + t ^{2}} + 1

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

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

\int 1 d t = t

= - arctan (t) + t

Füge eine Konstante zur Lösung hinzu

= - arctan (t) + t + C

\int 8 x^{3} + \frac{1}{x} d x

\frac{\partial}{\partial y} (e^{7 x e^{y}})

d er i v a t i v e f (x) = 6 x^{3} cot (x)

t an g e n t f (x) = x^{3} 5 - x^{2}, at x = 2

\frac{d}{d x} (- \frac{1}{2} e^{- x^{2}})