What is the integral of xln((1+2x)/(1-2x)) ?

The integral of xln((1+2x)/(1-2x)) is 1/8 (4x^2ln((2x+1)/(-2x+1))+4x-ln((2x+1)/(-2x+1))-2)+C

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

\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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\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

Expand

\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

Apply the Sum Rule:

\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})))

Substitute back

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

= \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})))

Simplify

\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)

Add a constant to the solution

= \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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What is the integral of xln((1+2x)/(1-2x)) ?
The integral of xln((1+2x)/(1-2x)) is 1/8 (4x^2ln((2x+1)/(-2x+1))+4x-ln((2x+1)/(-2x+1))-2)+C