integral from 1 to 2 of x/(x^2+4x+20)

Lösung

\int_{1}^{2} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (2) - ln (\frac{25}{16}) - \frac{π}{4} + arctan (\frac{3}{4}))

Dezimale

0.05248 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{x}{x ^{2} + 4 x + 20} d x

Vervollständige das Quadrat

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int_{1}^{2} \frac{x}{( x + 2 ) ^{2} + 16} d x

Wende U-Substitution an

= \int_{3}^{4} \frac{u - 2}{u ^{2} + 16} d u

Wende integrale Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{\frac{3}{4}}^{1} \frac{2 v - 1}{v ^{2} + 1} d v

Multipliziere aus

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

Wende die Summenregel an:

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

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

\int_{\frac{3}{4}}^{1} \frac{2 v}{v ^{2} + 1} d v = ln (2) - ln (\frac{25}{16})

\int_{\frac{3}{4}}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4} - arctan (\frac{3}{4})

= \frac{1}{2} (ln (2) - ln (\frac{25}{16}) - (\frac{π}{4} - arctan (\frac{3}{4})))

Vereinfache

= \frac{1}{2} (ln (2) - ln (\frac{25}{16}) - \frac{π}{4} + arctan (\frac{3}{4}))

\int_{0}^{0.4} \frac{( x ^{2} )}{4 - 25 x ^{2}} d x

\frac{\partial}{\partial x} (y x - y - x)

y^{^{'}} = 5 x - y

\frac{d}{d x} (x y^{2} - x^{3} y)

\frac{d}{d x} (x + \frac{1}{x ^{3}})