integral from-1 to 1 of (4x)/((x+3)^2)

Lösung

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

4 ln (2) - 3

Dezimale

- 0.22741 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{u - 3}{u ^{2}} : \frac{1}{u} - \frac{3}{u ^{2}}

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

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

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

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

= 4 (ln (2) - \frac{3}{4})

Vereinfache

4 (ln (2) - \frac{3}{4}) : 4 ln (2) - 3

= 4 ln (2) - 3

\int \frac{e ^{4 x}}{4} d x

x \to \infty lim (n^{x})

n = 1 \sum \infty \frac{n !}{n ^{2}}

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