integral of (x^2)/((x^2+16)^{3/2)}

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int tan^{2} (u) cos (u) d u

Vereinfache

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int \frac{1 - cos ^{2} ( u )}{cos ( u )} d u

Multipliziere aus

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

= \int \frac{1}{cos ( u )} - cos (u) 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{1}{cos ( u )} d u - \int cos (u) d u

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

\int cos (u) d u = sin (u)

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

Setze in

u = arctan (\frac{1}{4} x)

ein

= ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) - sin (arctan (\frac{1}{4} x))

Vereinfache

ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) - sin (arctan (\frac{1}{4} x)) : ln (\frac{1}{4} x + (16 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 16 + x ^{2} ) ^{\frac{1}{2}}}

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial x} (\frac{6 x}{y ^{2}})

in v erse l a pl a ce \frac{1}{10 \cdot s}

\int \frac{y}{3} d y

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

in v erse l a pl a ce \frac{2}{s ^{2}} \cdot e^{- s}