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

Lösung

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

\frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3)) + C

Schritte zur Lösung

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

Vereinfache

\frac{x}{( x - 3 ) ^{2} + 4 ^{2}} : \frac{x}{x ^{2} - 6 x + 25}

= \int \frac{x}{x ^{2} - 6 x + 25} d x

Vervollständige das Quadrat

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = x - 3

ein

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

Vereinfache

\frac{1}{2} ln (x - 3)^{2} + 16 + \frac{3}{4} arctan (\frac{x - 3}{4}) : \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3))

= \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 6 x + 25 + \frac{3}{4} arctan (\frac{1}{4} (x - 3)) + C

\int_{0.1}^{0.2} (300 x) d x

\frac{d}{d s} (\frac{2}{s ^{3}})

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d er i v a t i v e \frac{ln ( x )}{x ^{9}}

d er i v a t i v e 4 e^{6 x} + 2 x