integral from 6 to 7 of x/(x^2+4x+13)

Lösung

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 13} d x

\frac{ln ( 10 ) - ln ( \frac{73}{9} )}{2} - \frac{2}{3} arctan (3) + \frac{2}{3} arctan (\frac{8}{3})

Dezimale

0.07999 \dots

Schritte zur Lösung

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 13} d x

Vervollständige das Quadrat

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int_{6}^{7} \frac{x}{( x + 2 ) ^{2} + 9} d x

Wende U-Substitution an

= \int_{8}^{9} \frac{u - 2}{u ^{2} + 9} d u

Wende integrale Substitution an

= \int_{\frac{8}{3}}^{3} \frac{3 v - 2}{3 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{\frac{8}{3}}^{3} \frac{3 v - 2}{v ^{2} + 1} d v

Multipliziere aus

\frac{3 v - 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} - \frac{2}{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}{3} (\int_{\frac{8}{3}}^{3} \frac{3 v}{v ^{2} + 1} d v - \int_{\frac{8}{3}}^{3} \frac{2}{v ^{2} + 1} d v)

\int_{\frac{8}{3}}^{3} \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} (ln (10) - ln (\frac{73}{9}))

\int_{\frac{8}{3}}^{3} \frac{2}{v ^{2} + 1} d v = 2 (arctan (3) - arctan (\frac{8}{3}))

= \frac{1}{3} (\frac{3}{2} (ln (10) - ln (\frac{73}{9})) - 2 (arctan (3) - arctan (\frac{8}{3})))

Vereinfache

\frac{1}{3} (\frac{3}{2} (ln (10) - ln (\frac{73}{9})) - 2 (arctan (3) - arctan (\frac{8}{3}))) : \frac{ln ( 10 ) - ln ( \frac{73}{9} )}{2} - \frac{2}{3} arctan (3) + \frac{2}{3} arctan (\frac{8}{3})

= \frac{ln ( 10 ) - ln ( \frac{73}{9} )}{2} - \frac{2}{3} arctan (3) + \frac{2}{3} arctan (\frac{8}{3})

\frac{\partial}{\partial y} (7 (x + 5 y - 4)^{2})

\int_{- 2}^{2} (4 x^{3} + 6 x^{2} - 8 x + 5) d x

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\int_{0}^{1} \frac{2 x}{x ^{2} + 1} d x

in v erse l a pl a ce \frac{s}{1 + s}