What is the integral of ((2x^3-x))/(x^4-x^2-25) ?

The integral of ((2x^3-x))/(x^4-x^2-25) is 1/2 ln|x^4-x^2-25|+C

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integral of ((2x^3-x))/(x^4-x^2-25)

\int \frac{( 2 x ^{3} - x )}{x ^{4} - x ^{2} - 25} d x

\frac{1}{2} ln x^{4} - x^{2} - 25 + C

Solution steps

\int \frac{2 x ^{3} - x}{x ^{4} - x ^{2} - 25} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{4} - x^{2} - 25

= \frac{1}{2} ln x^{4} - x^{2} - 25

Add a constant to the solution

= \frac{1}{2} ln x^{4} - x^{2} - 25 + C

\frac{d}{d x} (\frac{x ^{3}}{( x - 2 ) ^{2}})

\frac{d}{d x} (\frac{tan ( x - 1 )}{x ^{3} + 2})

\int \frac{1}{( 2 x + 3 y ) ^{2}} d x

\int sin^{2} (3 x co) s^{2} 3 x d x

\frac{d y}{d x} = e^{y + x}

What is the integral of ((2x^3-x))/(x^4-x^2-25) ?
The integral of ((2x^3-x))/(x^4-x^2-25) is 1/2 ln|x^4-x^2-25|+C