integral of (2x^2)/((x^2+5)^2)

Lösung

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

\frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}} + C

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int \frac{x ^{2}}{( x ^{2} + 5 ) ^{2}} d x

Wende integrale Substitution an

= 2 \cdot \int \frac{u ^{2}}{5 ( u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{5} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Ermittle den Partialbruch von

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

= 2 \cdot \frac{1}{5} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} 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

= 2 \cdot \frac{1}{5} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= 2 \cdot \frac{1}{5} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Setze in

u = \frac{x}{5}

ein

= 2 \cdot \frac{1}{5} arctan (\frac{x}{5}) - \frac{1}{2} arctan (\frac{x}{5}) + \frac{\frac{x}{5}}{1 + ( \frac{x}{5} ) ^{2}}

Vereinfache

2 \cdot \frac{1}{5} arctan (\frac{x}{5}) - \frac{1}{2} arctan (\frac{x}{5}) + \frac{\frac{x}{5}}{1 + ( \frac{x}{5} ) ^{2}} : \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}}

= \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} arctan (\frac{x}{5}) - \frac{x}{5 + x ^{2}} + C

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

in t e g r a l cos^{2} (t)

\int \frac{1}{2}

\int sin (a x) e^{b x + c} d x

\int 5 sin (6 - 5 x) d x