integral of ((1+2arctan(x))^3)/(1+x^2)

Lösung

\int \frac{( 1 + 2 arctan ( x ) ) ^{3}}{1 + x ^{2}} d x

arctan (x) + 3 arctan^{2} (x) + 4 arctan^{3} (x) + 2 arctan^{4} (x) + C

Schritte zur Lösung

\int \frac{( 1 + 2 arctan ( x ) ) ^{3}}{1 + x ^{2}} d x

(1 + 2 arctan (x))^{3} = 1 + 6 arctan (x) + 12 arctan^{2} (x) + 8 arctan^{3} (x)

= \int \frac{1 + 6 arctan ( x ) + 12 arctan ^{2} ( x ) + 8 arctan ^{3} ( x )}{1 + x ^{2}} d x

Multipliziere aus

\frac{1 + 6 arctan ( x ) + 12 arctan ^{2} ( x ) + 8 arctan ^{3} ( x )}{1 + x ^{2}} : \frac{1}{1 + x ^{2}} + \frac{6 arctan ( x )}{1 + x ^{2}} + \frac{12 arctan ^{2} ( x )}{1 + x ^{2}} + \frac{8 arctan ^{3} ( x )}{1 + x ^{2}}

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{1}{1 + x ^{2}} d x + \int \frac{6 arctan ( x )}{1 + x ^{2}} d x + \int \frac{12 arctan ^{2} ( x )}{1 + x ^{2}} d x + \int \frac{8 arctan ^{3} ( x )}{1 + x ^{2}} d x

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

\int \frac{6 arctan ( x )}{1 + x ^{2}} d x = 3 arctan^{2} (x)

\int \frac{12 arctan ^{2} ( x )}{1 + x ^{2}} d x = 4 arctan^{3} (x)

\int \frac{8 arctan ^{3} ( x )}{1 + x ^{2}} d x = 2 arctan^{4} (x)

= arctan (x) + 3 arctan^{2} (x) + 4 arctan^{3} (x) + 2 arctan^{4} (x)

Füge eine Konstante zur Lösung hinzu

= arctan (x) + 3 arctan^{2} (x) + 4 arctan^{3} (x) + 2 arctan^{4} (x) + C

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