integral of 2/x+1/(4x^2+1)-x^{-3/2}

Lösung

\int \frac{2}{x} + \frac{1}{4 x ^{2} + 1} - x^{- \frac{3}{2}} d x

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

Schritte zur Lösung

\int \frac{2}{x} + \frac{1}{4 x ^{2} + 1} - x^{- \frac{3}{2}} d x

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} d x + \int \frac{1}{4 x ^{2} + 1} d x - \int x^{- \frac{3}{2}} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

\int x^{- \frac{3}{2}} d x = - \frac{2}{x ^{\frac{1}{2}}}

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int arctan (x) x^{2} d x

\int \frac{x ^{2} - x + 2}{x ^{2} + 1} d x

\int y 4 - y^{2} d y

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

\int \frac{1}{x ^{2} - a} d x