What is the integral of (x^2)/(sqrt(x^6+16)) ?

The integral of (x^2)/(sqrt(x^6+16)) is 1/3 ln((|x^3+sqrt(16+x^6)|)/4)+C

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integral of (x^2)/(sqrt(x^6+16))

\int \frac{x ^{2}}{x ^{6} + 16} d x

\frac{1}{3} ln (\frac{x ^{3} + 16 + x ^{6}}{4}) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} + 16} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 16} d u

= \int \frac{1}{3 u ^{2} + 16} 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}{u ^{2} + 16} 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}{4} x^{3})) + sec (arctan (\frac{1}{4} x^{3}))

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{4} x^{3})) + sec (arctan (\frac{1}{4} x^{3})) : \frac{1}{3} ln (\frac{x ^{3} + 16 + x ^{6}}{4})

= \frac{1}{3} ln (\frac{x ^{3} + 16 + x ^{6}}{4})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{3} + 16 + x ^{6}}{4}) + C

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What is the integral of (x^2)/(sqrt(x^6+16)) ?
The integral of (x^2)/(sqrt(x^6+16)) is 1/3 ln((|x^3+sqrt(16+x^6)|)/4)+C