What is the integral of x/(sqrt(x^4-2x^2+10)) ?

The integral of x/(sqrt(x^4-2x^2+10)) is 1/2 ln((|x^2-1+sqrt(x^4-2x^2+10)|)/3)+C

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integral of x/(sqrt(x^4-2x^2+10))

\int \frac{x}{x ^{4} - 2 x ^{2} + 10} d x

\frac{1}{2} ln (\frac{x ^{2} - 1 + x ^{4} - 2 x ^{2} + 10}{3}) + C

Solution steps

\int \frac{x}{x ^{4} - 2 x ^{2} + 10} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} - 2 u + 10} d u

= \int \frac{1}{2 u ^{2} - 2 u + 10} 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{1}{u ^{2} - 2 u + 10} d u

Complete the square

u^{2} - 2 u + 10 : (u - 1)^{2} + 9

= \frac{1}{2} \cdot \int \frac{1}{( u - 1 ) ^{2} + 9} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 9} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= \frac{1}{2} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (\frac{1}{3} (x^{2} - 1))) + sec (arctan (\frac{1}{3} (x^{2} - 1)))

Simplify

\frac{1}{2} ln tan (arctan (\frac{1}{3} (x^{2} - 1))) + sec (arctan (\frac{1}{3} (x^{2} - 1))) : \frac{1}{2} ln (\frac{x ^{2} - 1 + x ^{4} - 2 x ^{2} + 10}{3})

= \frac{1}{2} ln (\frac{x ^{2} - 1 + x ^{4} - 2 x ^{2} + 10}{3})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x ^{2} - 1 + x ^{4} - 2 x ^{2} + 10}{3}) + C

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What is the integral of x/(sqrt(x^4-2x^2+10)) ?
The integral of x/(sqrt(x^4-2x^2+10)) is 1/2 ln((|x^2-1+sqrt(x^4-2x^2+10)|)/3)+C