integral from 0 to 1 of (x+1)/(4+x^2)

Lösung

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

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

Dezimale

0.34339 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 1}{4 + x ^{2}} : \frac{x}{4 + x ^{2}} + \frac{1}{4 + x ^{2}}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{x}{4 + x ^{2}} d x + \int_{0}^{1} \frac{1}{4 + x ^{2}} d x

\int_{0}^{1} \frac{x}{4 + x ^{2}} d x = \frac{1}{2} ln (5) - ln (2)

\int_{0}^{1} \frac{1}{4 + x ^{2}} d x = \frac{1}{2} arctan (\frac{1}{2})

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

Vereinfache

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

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

\int_{0}^{3} - x^{2} - 3 x d x

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

\int_{0}^{6} 2 x d x

\int_{2}^{7} 1 d x

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