What is the integral of 1/((3x^2-4x)) ?

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

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integral of 1/((3x^2-4x))

integral

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

- \frac{1}{4} (ln (\frac{3}{2} ∣ x ∣) - ln \frac{3}{2} x - 2) + C

Solution steps

\int \frac{1}{3 x ^{2} - 4 x} d x

Complete the square

3 x^{2} - 4 x : 3 (x - \frac{2}{3})^{2} - \frac{4}{3}

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

Apply u-substitution

: \int \frac{3}{9 u ^{2} - 4} d u

= \int \frac{3}{9 u ^{2} - 4} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} - 4} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{6 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{6} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 3 \cdot \frac{1}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 3 \cdot \frac{1}{6} (- (\frac{ln \frac{3}{2} ( x - \frac{2}{3} ) + 1}{2} - \frac{ln \frac{3}{2} ( x - \frac{2}{3} ) - 1}{2}))

Simplify

3 \cdot \frac{1}{6} (- (\frac{ln \frac{3}{2} ( x - \frac{2}{3} ) + 1}{2} - \frac{ln \frac{3}{2} ( x - \frac{2}{3} ) - 1}{2})) : - \frac{1}{4} (ln (\frac{3}{2} ∣ x ∣) - ln \frac{3}{2} x - 2)

= - \frac{1}{4} (ln (\frac{3}{2} ∣ x ∣) - ln \frac{3}{2} x - 2)

Add a constant to the solution

= - \frac{1}{4} (ln (\frac{3}{2} ∣ x ∣) - ln \frac{3}{2} x - 2) + C

0.7 5^{x}

1 + y^{3}

\frac{1 + x + x ^{2}}{x ^{2} ( 1 + x )}

a + x^{2}

\frac{1}{cos ^{4} ( x )}

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