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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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