integral of ((x))/((4x^{(2))-12x+25)}

Lösung

\int \frac{( x )}{( 4 x ^{(2)} - 12 x + 25 )} d x

\frac{1}{16} (2 ln 4 x^{2} - 12 x + 25 + 3 arctan (\frac{2 x - 3}{4})) + C

Schritte zur Lösung

\int \frac{x}{4 x ^{2} - 12 x + 25} d x

Vervollständige das Quadrat

4 x^{2} - 12 x + 25 : 4 (x - \frac{3}{2})^{2} + 16

= \int \frac{x}{4 ( x - \frac{3}{2} ) ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{2 u + 3}{2 ( 4 u ^{2} + 16 )} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int \frac{2 u + 3}{4 u ^{2} + 16} d u

Multipliziere aus

\frac{2 u + 3}{4 u ^{2} + 16} : \frac{2 u}{4 u ^{2} + 16} + \frac{3}{4 u ^{2} + 16}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{2 u}{4 u ^{2} + 16} d u + \int \frac{3}{4 u ^{2} + 16} d u)

\int \frac{2 u}{4 u ^{2} + 16} d u = \frac{1}{4} ln 4 u^{2} + 16

\int \frac{3}{4 u ^{2} + 16} d u = \frac{3}{8} arctan (\frac{u}{2})

= \frac{1}{2} (\frac{1}{4} ln 4 u^{2} + 16 + \frac{3}{8} arctan (\frac{u}{2}))

Setze in

u = x - \frac{3}{2}

ein

= \frac{1}{2} (\frac{1}{4} ln 4 (x - \frac{3}{2})^{2} + 16 + \frac{3}{8} arctan (\frac{x - \frac{3}{2}}{2}))

Vereinfache

\frac{1}{2} (\frac{1}{4} ln 4 (x - \frac{3}{2})^{2} + 16 + \frac{3}{8} arctan (\frac{x - \frac{3}{2}}{2})) : \frac{1}{16} (2 ln 4 x^{2} - 12 x + 25 + 3 arctan (\frac{2 x - 3}{4}))

= \frac{1}{16} (2 ln 4 x^{2} - 12 x + 25 + 3 arctan (\frac{2 x - 3}{4}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (2 ln 4 x^{2} - 12 x + 25 + 3 arctan (\frac{2 x - 3}{4})) + C

x \to 4 lim (\frac{x - 5}{( x - 4 ) ^{2}})

x y^{^{'}} + 2 y = - 4 sin (x)

\int \frac{sec ^{4} ( t )}{tan ^{2} ( t )} d t

\frac{d}{d x} (\frac{x ^{2} + 2 x}{4 - 5 x})

t an g e n t f (x) = tan (x), at x = \frac{π}{4}