integral of x/(sqrt(x^2+8x))

Lösung

\int \frac{x}{x ^{2} + 8 x} d x

x^{2} + 8 x - 4 ln (\frac{1}{4} x^{2} + 8 x + x + 4) + C

Schritte zur Lösung

\int \frac{x}{x ^{2} + 8 x} d x

Vervollständige das Quadrat

x^{2} + 8 x : (x + 4)^{2} - 16

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

Wende U-Substitution an

= \int \frac{u - 4}{u ^{2} - 16} d u

Multipliziere aus

\frac{u - 4}{u ^{2} - 16} : \frac{u}{u ^{2} - 16} - \frac{4}{u ^{2} - 16}

= \int \frac{u}{u ^{2} - 16} - \frac{4}{u ^{2} - 16} d u

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{4}{u ^{2} - 16} d u

\int \frac{u}{u ^{2} - 16} d u = u^{2} - 16

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

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

Setze in

u = x + 4

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t y = a x^{3}

\int_{0}^{1} \frac{ln ( t )}{t} d t

\int + x^{3} e^{3 x^{2}} d x

x \to 1 + lim (\frac{x ^{2} - 4}{x - 1})

\int \frac{20000}{( x + 100 ) ^{3}} d x