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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{3 16 + 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}{16 + 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}{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

\int \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} d x

x \to + π lim (x cos (\frac{7 x}{6}))

\int 2 x yz e^{x^{2}} d x

\int \frac{4}{1 + t ^{2}} d t

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

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