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

Lösung

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

\frac{π}{4} - arctan (x)

Schritte zur Lösung

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

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

\frac{π}{4} - arctan (x)

= \frac{π}{4} - arctan (x)

\int_{0}^{1} x 27 x^{2} + 9 d x

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

\int_{\frac{- π}{2}}^{\frac{π}{2}} cos (2 x) d x

\int_{- 1}^{4} \frac{1}{x ^{2} + x - 2} d x

\int_{2}^{6} 3 d x