What is the integral of (3x^2-7)/(x^3-7x+6) ?

The integral of (3x^2-7)/(x^3-7x+6) is ln|x^3-7x+6|+C

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integral of (3x^2-7)/(x^3-7x+6)

\int \frac{3 x ^{2} - 7}{x ^{3} - 7 x + 6} d x

ln x^{3} - 7 x + 6 + C

Solution steps

\int \frac{3 x ^{2} - 7}{x ^{3} - 7 x + 6} 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 = x^{3} - 7 x + 6

= ln x^{3} - 7 x + 6

Add a constant to the solution

= ln x^{3} - 7 x + 6 + C

\int_{0}^{1} (t 4 + 9 t^{2}) d t

x + 2 y = x + 2 y = 8

y^{^{'}} = 5 x (x - 3)

\frac{d y}{d x} = \frac{1}{3} (x^{2} + 2)^{\frac{3}{2}}

\frac{\partial}{\partial x} (e^{y e^{x}})

What is the integral of (3x^2-7)/(x^3-7x+6) ?
The integral of (3x^2-7)/(x^3-7x+6) is ln|x^3-7x+6|+C