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

The integral of (2x)/(sqrt(x^4+4)) is ln(1/2 |x^2+sqrt(4+x^4)|)+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^{2} + 4 + x^{4}) + 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 (arctan (\frac{1}{2} x^{2})) + sec (arctan (\frac{1}{2} x^{2}))

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{6 x}{y}

x^{2} + 1

\int \frac{3 x}{x ^{2} + 6 x + 9} d x

\frac{d}{d x} (in (2 x))

x \to + 3 - lim (\frac{1}{9 - x ^{2}})

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