What is the integral of (35x^4-3)/(7x^5-3x) ?

The integral of (35x^4-3)/(7x^5-3x) is ln|7x^5-3x|+C

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integral of (35x^4-3)/(7x^5-3x)

\int \frac{35 x ^{4} - 3}{7 x ^{5} - 3 x} d x

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

Solution steps

\int \frac{35 x ^{4} - 3}{7 x ^{5} - 3 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 = 7 x^{5} - 3 x

= ln 7 x^{5} - 3 x

Add a constant to the solution

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

(- 7, - 3), (- 5, - 8)

\int \frac{x ^{2}}{2} + 3 x d x

\frac{d}{d x} ((x - 2) (x + 1) (x + 5))

\int (x^{2} - 4 x - 15) d x

\int x^{2} cos (4 x) d x

What is the integral of (35x^4-3)/(7x^5-3x) ?
The integral of (35x^4-3)/(7x^5-3x) is ln|7x^5-3x|+C