integral of x/((1+x^2)^{1/2)}

Lösung

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

(1 + x^{2})^{\frac{1}{2}} + C

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a ^{b}} = a^{- b}

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

= \int x (1 + x^{2})^{- \frac{1}{2}} d x

Wende U-Substitution an

= \int \frac{1}{2 u ^{\frac{1}{2}}} 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{1}{u ^{\frac{1}{2}}} d u

Wende die Potenzregel an

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Setze in

u = 1 + x^{2}

ein

= \frac{1}{2} \cdot 2 (1 + x^{2})^{\frac{1}{2}}

Vereinfache

\frac{1}{2} \cdot 2 (1 + x^{2})^{\frac{1}{2}} : (1 + x^{2})^{\frac{1}{2}}

= (1 + x^{2})^{\frac{1}{2}}

Füge eine Konstante zur Lösung hinzu

= (1 + x^{2})^{\frac{1}{2}} + C

\frac{d}{d x} (8 (0.9)^{x - 2})

x y^{^{'}} = y^{2} - y, y (2) = 1

\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{3}} d x

t an g e n t f (x) = 3 x^{2} - 5 x + 8, at x = 2

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