What is the integral of (10x^4-9x^2)/(2x^5-3x^3+14) ?

The integral of (10x^4-9x^2)/(2x^5-3x^3+14) is ln|2x^5-3x^3+14|+C

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integral of (10x^4-9x^2)/(2x^5-3x^3+14)

\int \frac{10 x ^{4} - 9 x ^{2}}{2 x ^{5} - 3 x ^{3} + 14} d x

ln 2 x^{5} - 3 x^{3} + 14 + C

Solution steps

\int \frac{10 x ^{4} - 9 x ^{2}}{2 x ^{5} - 3 x ^{3} + 14} 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 = 2 x^{5} - 3 x^{3} + 14

= ln 2 x^{5} - 3 x^{3} + 14

Add a constant to the solution

= ln 2 x^{5} - 3 x^{3} + 14 + C

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 0

\frac{\partial}{\partial x} (\frac{5 y}{x ^{3}})

y^{^{''}} - 9 y^{^{'}} + 18 y = 5 sin (4 t), y (0) = 5, y^{^{'}} (0) = 2

x e^{- 3 x}

\int e^{x} 81 - e^{2 x} d x

What is the integral of (10x^4-9x^2)/(2x^5-3x^3+14) ?
The integral of (10x^4-9x^2)/(2x^5-3x^3+14) is ln|2x^5-3x^3+14|+C