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

The integral of (-2x^3+x)/(-4x^4+4x^2) is 1/8 ln|-4x^4+4x^2|+C

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

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

\frac{1}{8} ln - 4 x^{4} + 4 x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

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

= \frac{1}{8} ln - 4 x^{4} + 4 x^{2}

Add a constant to the solution

= \frac{1}{8} ln - 4 x^{4} + 4 x^{2} + C

\int \frac{x}{1 - x ^{3}} d x

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

n = 1 \sum \infty n (n - 1) x^{n}

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

\int \frac{ln ( x )}{9} d x

What is the integral of (-2x^3+x)/(-4x^4+4x^2) ?
The integral of (-2x^3+x)/(-4x^4+4x^2) is 1/8 ln|-4x^4+4x^2|+C