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Beliebt Rechnen >

integral of (x^2)/(sqrt((1+4x)^5))

Lösung

\int \frac{x ^{2}}{( 1 + 4 x ) ^{5}} d x

Lösung

\frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}} + C

Schritte zur Lösung

\int \frac{x ^{2}}{( 1 + 4 x ) ^{5}} d x

(1 + 4 x)^{5} = (1 + 4 x)^{\frac{5}{2}},

angenommen

(1 + 4 x) \geq 0

= \int \frac{x ^{2}}{( 1 + 4 x ) ^{\frac{5}{2}}} d x

Wende U-Substitution an

= \int \frac{( u - 1 ) ^{2}}{64 u ^{\frac{5}{2}}} d u

Entferne die Konstante:

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

= \frac{1}{64} \cdot \int \frac{( u - 1 ) ^{2}}{u ^{\frac{5}{2}}} d u

Multipliziere aus

\frac{( u - 1 ) ^{2}}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{2}{u ^{\frac{3}{2}}} + \frac{1}{u ^{\frac{5}{2}}}

= \frac{1}{64} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{2}{u ^{\frac{3}{2}}} + \frac{1}{u ^{\frac{5}{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

= \frac{1}{64} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{2}{u ^{\frac{3}{2}}} d u + \int \frac{1}{u ^{\frac{5}{2}}} d u)

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

\int \frac{2}{u ^{\frac{3}{2}}} d u = - \frac{4}{u}

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

= \frac{1}{64} (2 u^{\frac{1}{2}} - (- \frac{4}{u}) - \frac{2}{3 u ^{\frac{3}{2}}})

Setze in

u = 1 + 4 x

ein

= \frac{1}{64} (2 (1 + 4 x)^{\frac{1}{2}} - (- \frac{4}{1 + 4 x}) - \frac{2}{3 ( 1 + 4 x ) ^{\frac{3}{2}}})

Vereinfache

\frac{1}{64} (2 (1 + 4 x)^{\frac{1}{2}} - (- \frac{4}{1 + 4 x}) - \frac{2}{3 ( 1 + 4 x ) ^{\frac{3}{2}}}) : \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}}

= \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}}

Füge eine Konstante zur Lösung hinzu

= \frac{6 x ^{2} + 6 x + 1}{12 ( 4 x + 1 ) ^{\frac{3}{2}}} + C

Graph

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