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

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

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

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

ln x^{9} + 8 x + C

Solution steps

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

= ln x^{9} + 8 x

Add a constant to the solution

= ln x^{9} + 8 x + C

\frac{\partial}{\partial x} (2 x e^{x^{2} + y^{2}})

8 y^{^{''}} + 32 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

\frac{x ^{2}}{2 + x}

\frac{d x}{d y} = (4 x + 2 y + 1)

y = (8 x^{2} - 16 x + 16) e^{x}

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