integral of x/(2x^2+2x+7)

Lösung

\int \frac{x}{2 x ^{2} + 2 x + 7} d x

\frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13}) + C

Schritte zur Lösung

\int \frac{x}{2 x ^{2} + 2 x + 7} d x

Vervollständige das Quadrat

2 x^{2} + 2 x + 7 : 2 (x + \frac{1}{2})^{2} + \frac{13}{2}

= \int \frac{x}{2 ( x + \frac{1}{2} ) ^{2} + \frac{13}{2}} d x

Wende U-Substitution an

= \int \frac{2 u - 1}{4 u ^{2} + 13} d u

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} + 13} : \frac{2 u}{4 u ^{2} + 13} - \frac{1}{4 u ^{2} + 13}

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 u}{4 u ^{2} + 13} d u - \int \frac{1}{4 u ^{2} + 13} d u

\int \frac{2 u}{4 u ^{2} + 13} d u = \frac{1}{4} ln 4 u^{2} + 13

\int \frac{1}{4 u ^{2} + 13} d u = \frac{1}{2 13} arctan (\frac{2}{13} u)

= \frac{1}{4} ln 4 u^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} u)

Setze in

u = x + \frac{1}{2}

ein

= \frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} (x + \frac{1}{2}))

Vereinfache

\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} (x + \frac{1}{2})) : \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13})

= \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13}) + C

y^{^{''}} + a^{2} y = sin (b \cdot t)

\int (6 x^{2} - 4 x + 2) (x^{3} - x^{2} + x + 1)^{7} d x

\frac{\partial}{\partial x} (a^{2} - x^{2} - y^{2})

\int \frac{e ^{x^{2}}}{x} d x

x \to 3 lim (\frac{5}{2 x - 6})