integral of xln((1+2x)/(1-2x))

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = \frac{1 + 2 x}{1 - 2 x}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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