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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplify

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

= \frac{1}{4} \cdot \int \frac{1}{u ( u - 1 )} d u

Apply u-substitution

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

Take the constant out:

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

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{4} \cdot 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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\int x 9 - 4 x^{2} d x

\int \frac{x ^{8}}{1 + x ^{3}} d x

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