What is the integral of (4x-3)/(4x^2-6x) ?

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

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integral of (4x-3)/(4x^2-6x)

\int \frac{4 x - 3}{4 x ^{2} - 6 x} d x

\frac{1}{2} ln 4 x^{2} - 6 x + C

Solution steps

\int \frac{4 x - 3}{4 x ^{2} - 6 x} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 4 x^{2} - 6 x

= \frac{1}{2} ln 4 x^{2} - 6 x

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} - 6 x + C

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

\int tanh (\frac{x}{6}) d x

x \to 0 lim (2 cos (\frac{1}{x}))

3 cos (2 x)

y = x^{3}, at x = - 3

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