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

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

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

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

- \frac{1}{4 ( x ^{4} - 2 x ^{2} + 3 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u ^{2}} d u

= \int \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{4} (- \frac{1}{u})

Substitute back

u = x^{4} - 2 x^{2} + 3

= \frac{1}{4} (- \frac{1}{x ^{4} - 2 x ^{2} + 3})

Simplify

\frac{1}{4} (- \frac{1}{x ^{4} - 2 x ^{2} + 3}) : - \frac{1}{4 ( x ^{4} - 2 x ^{2} + 3 )}

= - \frac{1}{4 ( x ^{4} - 2 x ^{2} + 3 )}

Add a constant to the solution

= - \frac{1}{4 ( x ^{4} - 2 x ^{2} + 3 )} + C

\int 2 t^{- 4} d t

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y^{^{''}} + y = 3 sin (2 t) + 2 t (cos (2 t))

\int x^{3} 49 - x^{2} d x

\frac{d}{d x} (a (2 x e^{3 x} + e^{3 x} \cdot 3 x^{2}))

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