integral of (2t^2-1)/((t^2+1)^2)

Lösung

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

- \frac{t}{2 ( t ^{2} + 1 )} + \frac{1}{2} arctan (t) - \frac{t}{t ^{2} + 1} + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

= arctan (t) - \frac{t}{1 + t ^{2}} - \frac{1}{2} (arctan (t) + \frac{t}{1 + t ^{2}})

Vereinfache

arctan (t) - \frac{t}{1 + t ^{2}} - \frac{1}{2} (arctan (t) + \frac{t}{1 + t ^{2}}) : - \frac{t}{2 ( t ^{2} + 1 )} + \frac{1}{2} arctan (t) - \frac{t}{t ^{2} + 1}

= - \frac{t}{2 ( t ^{2} + 1 )} + \frac{1}{2} arctan (t) - \frac{t}{t ^{2} + 1}

Füge eine Konstante zur Lösung hinzu

= - \frac{t}{2 ( t ^{2} + 1 )} + \frac{1}{2} arctan (t) - \frac{t}{t ^{2} + 1} + C

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