integral of 3ln(x^2-x+4)

Lösung

\int 3 ln (x^{2} - x + 4) d x

3 (x ln (x^{2} - x + 4) - \frac{1}{2} ln (x^{2} - x + 4) + 15 arctan (\frac{1}{15} (2 x - 1)) - 2 x + 1) + C

Schritte zur Lösung

\int 3 ln (x^{2} - x + 4) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int ln (x^{2} - x + 4) d x

Vervollständige das Quadrat

x^{2} - x + 4 : (x - \frac{1}{2})^{2} + \frac{15}{4}

= 3 \cdot \int ln ((x - \frac{1}{2})^{2} + \frac{15}{4}) d x

Wende U-Substitution an

= 3 \cdot \int ln (u^{2} + \frac{15}{4}) d u

Wende die partielle Integration an

= 3 (u ln (u^{2} + \frac{15}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 15} d u)

\int \frac{8 u ^{2}}{4 u ^{2} + 15} d u = - 15 arctan (\frac{2}{15} u) + 2 u

= 3 (u ln (u^{2} + \frac{15}{4}) - (- 15 arctan (\frac{2}{15} u) + 2 u))

Setze in

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

ein

= 3 ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{15}{4}) - (- 15 arctan (\frac{2}{15} (x - \frac{1}{2})) + 2 (x - \frac{1}{2})))

Vereinfache

3 ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{15}{4}) - (- 15 arctan (\frac{2}{15} (x - \frac{1}{2})) + 2 (x - \frac{1}{2}))) : 3 (x ln (x^{2} - x + 4) - \frac{1}{2} ln (x^{2} - x + 4) + 15 arctan (\frac{1}{15} (2 x - 1)) - 2 x + 1)

= 3 (x ln (x^{2} - x + 4) - \frac{1}{2} ln (x^{2} - x + 4) + 15 arctan (\frac{1}{15} (2 x - 1)) - 2 x + 1)

Füge eine Konstante zur Lösung hinzu

= 3 (x ln (x^{2} - x + 4) - \frac{1}{2} ln (x^{2} - x + 4) + 15 arctan (\frac{1}{15} (2 x - 1)) - 2 x + 1) + C

((x - 2)^{2} (x + 4))^{^{'}}

f (x) = 5 x^{2} e^{3 x}

t an g e n t y = (1 + 2 x)^{2}, (3, 49)

in v erse l a pl a ce \frac{s}{( s + 1 ) ^{2} + 4}

\int 5 cos^{3} (2 x) d x