integral from 1 to 3 of (2x+3)/(x+1)

Lösung

\int_{1}^{3} \frac{2 x + 3}{x + 1} d x

4 + ln (2)

Dezimale

4.69314 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{2 x + 3}{x + 1} d x

Wende U-Substitution an

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

Multipliziere aus

\frac{2 u + 1}{u} : 2 + \frac{1}{u}

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

Wende die Summenregel an:

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

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

\int_{2}^{4} 2 d u = 4

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= 4 + ln (2)

\int_{- 3}^{4} (x^{2} + 3 x + 1) d x

\int_{1}^{4} \frac{( ln ( x ) + 4 ) ^{4}}{x} d x

\int_{1}^{2} \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

\int_{2}^{k} x - 2 d x

\int_{0}^{2} (x^{4} + 5)^{3} x^{3} d x