What is the integral of (9x^8+6)/(x^9+6x) ?

The integral of (9x^8+6)/(x^9+6x) is ln|x^9+6x|+C

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integral of (9x^8+6)/(x^9+6x)

\int \frac{9 x ^{8} + 6}{x ^{9} + 6 x} d x

ln x^{9} + 6 x + C

Solution steps

\int \frac{9 x ^{8} + 6}{x ^{9} + 6 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 = x^{9} + 6 x

= ln x^{9} + 6 x

Add a constant to the solution

= ln x^{9} + 6 x + C

f (x) = sin (7 ln (x))

\frac{d}{d x} (cos (3 cos (2 x)))

e^{- 2 s}

f^{^{'}} (x) = - \frac{1}{x} (f (x))

\frac{s + 1}{s ^{2} + 2 s + 2}

What is the integral of (9x^8+6)/(x^9+6x) ?
The integral of (9x^8+6)/(x^9+6x) is ln|x^9+6x|+C