What is the integral of (20x^3+8)/(5x^4+8x+3) ?

The integral of (20x^3+8)/(5x^4+8x+3) is ln|5x^4+8x+3|+C

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integral of (20x^3+8)/(5x^4+8x+3)

\int \frac{20 x ^{3} + 8}{5 x ^{4} + 8 x + 3} d x

ln 5 x^{4} + 8 x + 3 + C

Solution steps

\int \frac{20 x ^{3} + 8}{5 x ^{4} + 8 x + 3} 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 = 5 x^{4} + 8 x + 3

= ln 5 x^{4} + 8 x + 3

Add a constant to the solution

= ln 5 x^{4} + 8 x + 3 + C

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What is the integral of (20x^3+8)/(5x^4+8x+3) ?
The integral of (20x^3+8)/(5x^4+8x+3) is ln|5x^4+8x+3|+C