integral of (2x+1)/((x+1)^2(x^2+1))

Lösung

\int \frac{2 x + 1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x

\frac{1}{x + 1} + arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln x^{2} + 1 + C

Schritte zur Lösung

\int \frac{2 x + 1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x

Multipliziere aus

\frac{2 x + 1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} : \frac{2 x}{( x + 1 ) ^{2} ( x ^{2} + 1 )} + \frac{1}{( x + 1 ) ^{2} ( x ^{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{2 x}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x + \int \frac{1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x

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

\int \frac{1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln x^{2} + 1

= \frac{1}{x + 1} + arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln x^{2} + 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{x + 1} + arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{4} ln x^{2} + 1 + C

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