What is the integral of x/(sqrt(x^4-6x^2+5)) ?

The integral of x/(sqrt(x^4-6x^2+5)) is 1/2 ln|x^2+sqrt(x^4-1)|+C

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integral of x/(sqrt(x^4-6x^2+5))

\int \frac{x}{x ^{4} - 6 x ^{2} + 5} d x

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

Solution steps

\int \frac{x}{x ^{4} - 6 x ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{4 u u + 5 - 6 u} d u

= \int \frac{1}{4 u u + 5 - 6 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 + 5 - 6 u} d u

Apply exponent rule:

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

\frac{1}{u u + 5 - 6 u} = u^{- \frac{1}{2}} \frac{1}{u + 5 - 6}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 5 - 6} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{u + 5 - 6} : \frac{1}{u u - 1}

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

Apply u-substitution

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

= \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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of x/(sqrt(x^4-6x^2+5)) ?
The integral of x/(sqrt(x^4-6x^2+5)) is 1/2 ln|x^2+sqrt(x^4-1)|+C