integral from 1 to 2 of arctan(x)

Lösung

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

2 arctan (2) - \frac{1}{2} ln (5) - \frac{π}{4} + \frac{1}{2} ln (2)

Dezimale

0.97075 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{1}^{2}

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{1}^{2}

Berechne die Grenzen:

2 arctan (2) - \frac{1}{2} ln (5) - \frac{π}{4} + \frac{1}{2} ln (2)

= 2 arctan (2) - \frac{1}{2} ln (5) - \frac{π}{4} + \frac{1}{2} ln (2)

\int_{0}^{0.8} 48000 x - 21600 d x

\int_{0}^{1} (3 x)^{2} - (3 x^{2})^{2} d x

\int_{7}^{x} t^{2} + 15 d t

\int_{1}^{2} (1 + cos (x))^{2} d x

\int_{0}^{\frac{7}{4}} \frac{1}{16 x ^{2} + 49} d x