integral from 0 to x of (2t-3)/(4t^2+7)

Lösung

\int_{0}^{x} \frac{2 t - 3}{4 t ^{2} + 7} d t

\frac{1}{4} (ln 4 x^{2} + 7 - ln (7)) - \frac{3}{2 7} arctan (\frac{2 x}{7})

Schritte zur Lösung

\int_{0}^{x} \frac{2 t - 3}{4 t ^{2} + 7} d t

Multipliziere aus

\frac{2 t - 3}{4 t ^{2} + 7} : \frac{2 t}{4 t ^{2} + 7} - \frac{3}{4 t ^{2} + 7}

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}^{x} \frac{2 t}{4 t ^{2} + 7} d t - \int_{0}^{x} \frac{3}{4 t ^{2} + 7} d t

\int_{0}^{x} \frac{2 t}{4 t ^{2} + 7} d t = \frac{1}{4} (ln 4 x^{2} + 7 - ln (7))

\int_{0}^{x} \frac{3}{4 t ^{2} + 7} d t = \frac{3}{2 7} arctan (\frac{2 x}{7})

= \frac{1}{4} (ln 4 x^{2} + 7 - ln (7)) - \frac{3}{2 7} arctan (\frac{2 x}{7})

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