integral from-1 to 1 of (5x)/((x+4)^2)

Lösung

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

5 (ln (5) - ln (3) - \frac{8}{15})

Dezimale

- 0.11253 \dots

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Wende U-Substitution an

= 5 \cdot \int_{3}^{5} \frac{u - 4}{u ^{2}} d u

Multipliziere aus

\frac{u - 4}{u ^{2}} : \frac{1}{u} - \frac{4}{u ^{2}}

= 5 \cdot \int_{3}^{5} \frac{1}{u} - \frac{4}{u ^{2}} 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

= 5 (\int_{3}^{5} \frac{1}{u} d u - \int_{3}^{5} \frac{4}{u ^{2}} d u)

\int_{3}^{5} \frac{1}{u} d u = ln (5) - ln (3)

\int_{3}^{5} \frac{4}{u ^{2}} d u = \frac{8}{15}

= 5 (ln (5) - ln (3) - \frac{8}{15})

a re a y = ∣ x - 6 ∣, y = \frac{x}{2}

n = 1 \sum \infty \frac{2 n}{5 n - 3}

d er i v a t i v e e^{s i n (2 x)}

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

\int \frac{4 x}{4 x ^{2} + 1} d x