What is the integral of (21x^2-10x)/(7x^3-5x^2+13) ?

The integral of (21x^2-10x)/(7x^3-5x^2+13) is ln|7x^3-5x^2+13|+C

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integral of (21x^2-10x)/(7x^3-5x^2+13)

\int \frac{21 x ^{2} - 10 x}{7 x ^{3} - 5 x ^{2} + 13} d x

ln 7 x^{3} - 5 x^{2} + 13 + C

Solution steps

\int \frac{21 x ^{2} - 10 x}{7 x ^{3} - 5 x ^{2} + 13} 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 = 7 x^{3} - 5 x^{2} + 13

= ln 7 x^{3} - 5 x^{2} + 13

Add a constant to the solution

= ln 7 x^{3} - 5 x^{2} + 13 + C

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What is the integral of (21x^2-10x)/(7x^3-5x^2+13) ?
The integral of (21x^2-10x)/(7x^3-5x^2+13) is ln|7x^3-5x^2+13|+C