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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

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

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

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arcsec (\frac{1}{2} x^{2})) + sec (arcsec (\frac{1}{2} x^{2}))

Simplify

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

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

Add a constant to the solution

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

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\int \frac{x ^{3}}{( x ^{4} - 5 ) ^{7}} d x

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