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

The integral of (4x^3-3x)/(x^4-2x^2+5) is 1/4 (4ln|x^4-2x^2+5|+arctan((x^2-1)/2))+C

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

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

\frac{1}{4} (4 ln x^{4} - 2 x^{2} + 5 + arctan (\frac{x ^{2} - 1}{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{4 u - 3}{2 ( u ^{2} + 5 - 2 u )} d u

= \int \frac{4 u - 3}{2 ( u ^{2} + 5 - 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{4 u - 3}{u ^{2} + 5 - 2 u} d u

Complete the square

u^{2} + 5 - 2 u : (u - 1)^{2} + 4

= \frac{1}{2} \cdot \int \frac{4 u - 3}{( u - 1 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{4 v + 1}{v ^{2} + 4} d v

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

Expand

\frac{4 v + 1}{v ^{2} + 4} : \frac{4 v}{v ^{2} + 4} + \frac{1}{v ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{4 v}{v ^{2} + 4} d v + \int \frac{1}{v ^{2} + 4} d v)

\int \frac{4 v}{v ^{2} + 4} d v = 2 ln v^{2} + 4

\int \frac{1}{v ^{2} + 4} d v = \frac{1}{2} arctan (\frac{v}{2})

= \frac{1}{2} (2 ln v^{2} + 4 + \frac{1}{2} arctan (\frac{v}{2}))

Substitute back

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

Simplify

\frac{1}{2} (2 ln (x^{2} - 1)^{2} + 4 + \frac{1}{2} arctan (\frac{x ^{2} - 1}{2})) : \frac{1}{4} (4 ln x^{4} - 2 x^{2} + 5 + arctan (\frac{x ^{2} - 1}{2}))

= \frac{1}{4} (4 ln x^{4} - 2 x^{2} + 5 + arctan (\frac{x ^{2} - 1}{2}))

Add a constant to the solution

= \frac{1}{4} (4 ln x^{4} - 2 x^{2} + 5 + arctan (\frac{x ^{2} - 1}{2})) + C

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What is the integral of (4x^3-3x)/(x^4-2x^2+5) ?
The integral of (4x^3-3x)/(x^4-2x^2+5) is 1/4 (4ln|x^4-2x^2+5|+arctan((x^2-1)/2))+C