What is the integral of (x^2)/(sqrt(4+x^6)) ?

The integral of (x^2)/(sqrt(4+x^6)) is 1/3 ln((|x^3+sqrt(4+x^6)|)/2)+C

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integral of (x^2)/(sqrt(4+x^6))

\int \frac{x ^{2}}{4 + x ^{6}} d x

\frac{1}{3} ln (\frac{x ^{3} + 4 + x ^{6}}{2}) + C

Solution steps

\int \frac{x ^{2}}{4 + x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 4 + u ^{2}} d u

= \int \frac{1}{3 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}{3} \cdot \int \frac{1}{4 + u ^{2}} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{2} x^{3})) + sec (arctan (\frac{1}{2} x^{3})) : \frac{1}{3} ln (\frac{x ^{3} + 4 + x ^{6}}{2})

= \frac{1}{3} ln (\frac{x ^{3} + 4 + x ^{6}}{2})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{3} + 4 + x ^{6}}{2}) + C

x^{^{''}} + 3 x = 0

\frac{\partial}{\partial x} (x^{2} e^{- y})

\int 2 cos (x) - \frac{3}{1 - x ^{2}} d x

f (x) = 2 x^{2}, at x = 3

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

What is the integral of (x^2)/(sqrt(4+x^6)) ?
The integral of (x^2)/(sqrt(4+x^6)) is 1/3 ln((|x^3+sqrt(4+x^6)|)/2)+C