integral of (x+3)/(3x^2-6x+8)

Lösung

\int \frac{x + 3}{3 x ^{2} - 6 x + 8} d x

\frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1)) + C

Schritte zur Lösung

\int \frac{x + 3}{3 x ^{2} - 6 x + 8} d x

Multipliziere aus

\frac{x + 3}{3 x ^{2} - 6 x + 8} : \frac{x}{3 x ^{2} - 6 x + 8} + \frac{3}{3 x ^{2} - 6 x + 8}

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{x}{3 x ^{2} - 6 x + 8} d x + \int \frac{3}{3 x ^{2} - 6 x + 8} d x

\int \frac{x}{3 x ^{2} - 6 x + 8} d x = \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1))

\int \frac{3}{3 x ^{2} - 6 x + 8} d x = \frac{3}{5} arctan (\frac{3}{5} (x - 1))

= \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1)) + C

\frac{d ^{2}}{d x ^{2}} (4 x 3 x)

\int_{1}^{3} (x^{2} + 2 x - 4) d x

n = 1 \sum \infty \frac{5 ^{n}}{3 ^{n}}

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

3 y^{^{''}} - 24 y^{^{'}} + 36 y = 6 sin (2 x)