What is the integral of (8x^3+6x-5)/(2x^4+3x^2-5x) ?

The integral of (8x^3+6x-5)/(2x^4+3x^2-5x) is ln|2x^4+3x^2-5x|+C

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integral of (8x^3+6x-5)/(2x^4+3x^2-5x)

\int \frac{8 x ^{3} + 6 x - 5}{2 x ^{4} + 3 x ^{2} - 5 x} d x

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

Solution steps

\int \frac{8 x ^{3} + 6 x - 5}{2 x ^{4} + 3 x ^{2} - 5 x} 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^{4} + 3 x^{2} - 5 x

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

Add a constant to the solution

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

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What is the integral of (8x^3+6x-5)/(2x^4+3x^2-5x) ?
The integral of (8x^3+6x-5)/(2x^4+3x^2-5x) is ln|2x^4+3x^2-5x|+C