What is the integral of (6x-1)/((2x-1)x^3) ?

The integral of (6x-1)/((2x-1)x^3) is 4(2ln|2x-1|-2ln(2|x|)+1/x-1/(8x^2))+C

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integral of (6x-1)/((2x-1)x^3)

\int \frac{6 x - 1}{( 2 x - 1 ) x ^{3}} d x

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

Solution steps

\int \frac{6 x - 1}{( 2 x - 1 ) x ^{3}} d x

Apply u-substitution

= \int \frac{4 ( 3 u + 2 )}{u ( u + 1 ) ^{3}} d u

Take the constant out:

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

= 4 \cdot \int \frac{3 u + 2}{u ( u + 1 ) ^{3}} d u

Take the partial fraction of

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

= 4 \cdot \int \frac{2}{u} - \frac{2}{u + 1} - \frac{2}{( u + 1 ) ^{2}} + \frac{1}{( 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

= 4 (\int \frac{2}{u} d u - \int \frac{2}{u + 1} d u - \int \frac{2}{( u + 1 ) ^{2}} d u + \int \frac{1}{( u + 1 ) ^{3}} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

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

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

= 4 (2 ln ∣ u ∣ - 2 ln ∣ u + 1 ∣ - (- \frac{2}{u + 1}) - \frac{1}{2 ( u + 1 ) ^{2}})

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (6x-1)/((2x-1)x^3) ?
The integral of (6x-1)/((2x-1)x^3) is 4(2ln|2x-1|-2ln(2|x|)+1/x-1/(8x^2))+C