What is the integral of (2x)/(sqrt(x^4+16)) ?

The integral of (2x)/(sqrt(x^4+16)) is ln(1/4 |x^2+sqrt(16+x^4)|)+C

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integral of (2x)/(sqrt(x^4+16))

\int \frac{2 x}{x ^{4} + 16} d x

ln (\frac{1}{4} x^{2} + 16 + x^{4}) + C

Solution steps

\int \frac{2 x}{x ^{4} + 16} 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} + 16} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 16} d u

= 2 \cdot \int \frac{1}{2 u ^{2} + 16} 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} + 16} 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}{4} x^{2})) + sec (arctan (\frac{1}{4} x^{2}))

Simplify

2 \cdot \frac{1}{2} ln tan (arctan (\frac{1}{4} x^{2})) + sec (arctan (\frac{1}{4} x^{2})) : ln (\frac{1}{4} x^{2} + 16 + x^{4})

= ln (\frac{1}{4} x^{2} + 16 + x^{4})

Add a constant to the solution

= ln (\frac{1}{4} x^{2} + 16 + x^{4}) + C

3 cos (2 x), 3 - 3 cos (2 x)

x^{2} \cdot y^{^{'}} + (2 - y) = 0

\frac{\partial}{\partial x} ((2 x + 3 y)^{10})

\frac{d}{d x} (e^{x} \cdot sin (x))

x \to 0 lim (2 x^{2} - x)

What is the integral of (2x)/(sqrt(x^4+16)) ?
The integral of (2x)/(sqrt(x^4+16)) is ln(1/4 |x^2+sqrt(16+x^4)|)+C