integral ((x+2))/((x^2+2))

Lösung

integral

\frac{( x + 2 )}{( x ^{2} + 2 )}

\frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2}) + C

Schritte zur Lösung

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

Wende integrale Substitution an

= \int \frac{u + 2}{u ^{2} + 1} d u

Multipliziere aus

\frac{u + 2}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{2}{u ^{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 \frac{u}{u ^{2} + 1} d u + \int \frac{2}{u ^{2} + 1} d u

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

= \frac{1}{2} ln u^{2} + 1 + 2 arctan (u)

Setze in

u = \frac{x}{2}

ein

= \frac{1}{2} ln (\frac{x}{2})^{2} + 1 + 2 arctan (\frac{x}{2})

(\frac{x}{2})^{2} = \frac{x ^{2}}{2}

= \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{x ^{2}}{2} + 1 + 2 arctan (\frac{x}{2}) + C

in t e g r a l \frac{( z + 4 )}{( z ^{2} + 2 z + 5 )}

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in t e g r a l \frac{y ^{2}}{2}