What is the integral of (5x^4+6x)/(x^5+3x^2+7) ?

The integral of (5x^4+6x)/(x^5+3x^2+7) is ln|x^5+3x^2+7|+C

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integral of (5x^4+6x)/(x^5+3x^2+7)

\int \frac{5 x ^{4} + 6 x}{x ^{5} + 3 x ^{2} + 7} d x

ln x^{5} + 3 x^{2} + 7 + C

Solution steps

\int \frac{5 x ^{4} + 6 x}{x ^{5} + 3 x ^{2} + 7} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{5} + 3 x^{2} + 7

= ln x^{5} + 3 x^{2} + 7

Add a constant to the solution

= ln x^{5} + 3 x^{2} + 7 + C

What is the integral of (5x^4+6x)/(x^5+3x^2+7) ?
The integral of (5x^4+6x)/(x^5+3x^2+7) is ln|x^5+3x^2+7|+C