integral from 1 to 3 of (y+4)/(y-4)

Lösung

\int_{1}^{3} \frac{y + 4}{y - 4} d y

- 8 ln (3) + 2

Dezimale

- 6.78889 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{y + 4}{y - 4} d y

\frac{y + 4}{y - 4} = \frac{8}{y - 4} + 1

= \int_{1}^{3} \frac{8}{y - 4} + 1 d y

Wende die Summenregel an:

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

= \int_{1}^{3} \frac{8}{y - 4} d y + \int_{1}^{3} 1 d y

\int_{1}^{3} \frac{8}{y - 4} d y = - 8 ln (3)

\int_{1}^{3} 1 d y = 2

= - 8 ln (3) + 2

\int_{1}^{6} x^{3} + x^{2} + 3 d x

\int_{1}^{\infty} \frac{x}{x ^{2} + 7} d x

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

\int_{- 2}^{4} \frac{x + 1}{x + 6} d x

\int_{0}^{1} x ln (x^{2} + 1) d x