What is the integral of (4x-3)/(2x^2-3x-14) ?

The integral of (4x-3)/(2x^2-3x-14) is ln|2x^2-3x-14|+C

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integral of (4x-3)/(2x^2-3x-14)

\int \frac{4 x - 3}{2 x ^{2} - 3 x - 14} d x

ln 2 x^{2} - 3 x - 14 + C

Solution steps

\int \frac{4 x - 3}{2 x ^{2} - 3 x - 14} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 2 x^{2} - 3 x - 14

= ln 2 x^{2} - 3 x - 14

Add a constant to the solution

= ln 2 x^{2} - 3 x - 14 + C

y^{^{'''}} + 8 y^{^{''}} + 5 y^{^{'}} - 50 y = 0

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

\frac{d}{d x} (e^{x t a n (x)})

y = x^{4} + 3 x^{2} - x

y^{2} = x - 1, 1, 3

What is the integral of (4x-3)/(2x^2-3x-14) ?
The integral of (4x-3)/(2x^2-3x-14) is ln|2x^2-3x-14|+C