integral from-1 to 1 of (x+3)/(x^2+2x+5)

Lösung

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

\frac{1}{2} ln (2) + \frac{π}{4}

Dezimale

1.13197 \dots

Schritte zur Lösung

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

Vervollständige das Quadrat

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Wende U-Substitution an

= \int_{0}^{2} \frac{u + 2}{u ^{2} + 4} d u

Wende integrale Substitution an

= \int_{0}^{1} \frac{v + 1}{v ^{2} + 1} d v

Multipliziere aus

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

= \int_{0}^{1} \frac{v}{v ^{2} + 1} d v + \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

\int_{0}^{1} \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln (2)

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

= \frac{1}{2} ln (2) + \frac{π}{4}

\int_{0}^{1} \frac{x}{x ^{2} - 3 x} d x

\int_{0}^{t} e^{- u} (t - u)^{2} d u

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\int_{- 2}^{0} 3 x^{2} - 2 d x

\int_{0}^{5} 1 + t^{2} d t