What is the integral of (12x^3+6x^2)/(3x^4+2x^3-5) ?

The integral of (12x^3+6x^2)/(3x^4+2x^3-5) is ln|3x^4+2x^3-5|+C

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integral of (12x^3+6x^2)/(3x^4+2x^3-5)

\int \frac{12 x ^{3} + 6 x ^{2}}{3 x ^{4} + 2 x ^{3} - 5} d x

ln 3 x^{4} + 2 x^{3} - 5 + C

Solution steps

\int \frac{12 x ^{3} + 6 x ^{2}}{3 x ^{4} + 2 x ^{3} - 5} 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 = 3 x^{4} + 2 x^{3} - 5

= ln 3 x^{4} + 2 x^{3} - 5

Add a constant to the solution

= ln 3 x^{4} + 2 x^{3} - 5 + C

\frac{\partial}{\partial t} (1 + t)

f (x) = 4 + 7 x

\frac{d y}{d x} + \frac{2}{x} y = 4 x

\int_{3}^{8} \frac{1}{x ^{2} - 1} d x

What is the integral of (12x^3+6x^2)/(3x^4+2x^3-5) ?
The integral of (12x^3+6x^2)/(3x^4+2x^3-5) is ln|3x^4+2x^3-5|+C