integral from 1 to 2x of (t^2)/(t^2+1)

Lösung

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

- arctan (2 x) + \frac{π}{4} + 2 x - 1

Schritte zur Lösung

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

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

= \int_{1}^{2 x} - \frac{1}{t ^{2} + 1} + 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_{1}^{2 x} \frac{1}{t ^{2} + 1} d t + \int_{1}^{2 x} 1 d t

\int_{1}^{2 x} \frac{1}{t ^{2} + 1} d t = arctan (2 x) - \frac{π}{4}

\int_{1}^{2 x} 1 d t = 2 x - 1

= - (arctan (2 x) - \frac{π}{4}) + 2 x - 1

Vereinfache

= - arctan (2 x) + \frac{π}{4} + 2 x - 1

x \to - \infty lim (9 e^{2 x} - 7 e^{- 14 x} - e^{x})

t an g e n t f (x) = \frac{2}{4 - x}, at x = - 5

n = 1 \sum \infty \frac{n}{5 n ^{2} - 1}

\frac{\partial}{\partial x} (e^{5 x + 8 y})

\frac{d}{d x} (4 e^{x} x^{3 x})