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

Lösung

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

arctan (x) - \frac{π}{4}

Schritte zur Lösung

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

Nutze das gemeinsame Integral :

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

= [arctan (t)]_{1}^{x}

Berechne die Grenzen:

arctan (x) - \frac{π}{4}

= arctan (x) - \frac{π}{4}

d er i v a t i v e e^{x} (3 x^{2} - 5 x)

\frac{d y}{d x} = \frac{( - y ^{2} )}{( 1 + y ^{2} )} \cdot (e^{- x})

\int \frac{e ^{3 x}}{x} d x

t an g e n t f (x) = \frac{x}{x + 1}

x \to - 2 lim (\frac{1}{( x + 2 ) ^{5}})