integral from 4 to x of 1/(1+t^2)

Lösung

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

arctan (x) - arctan (4)

Schritte zur Lösung

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

Nutze das gemeinsame Integral :

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

= [arctan (t)]_{4}^{x}

Berechne die Grenzen:

arctan (x) - arctan (4)

= arctan (x) - arctan (4)

\int_{0}^{1} x^{4} - x^{2} d x

\int_{0}^{\frac{π}{4}} \frac{d}{1} (cos^{4} (x)) d x

\int_{2}^{6} (x^{2} + 7 x - 5) (x - 2) d x

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

\int_{- \frac{π}{4}}^{\frac{π}{4}} sin^{2} (x) d x