What is the integral of (2x^2)/(x^3sqrt(x^6-9)) ?

The integral of (2x^2)/(x^3sqrt(x^6-9)) is 2/9 arctan(1/3 sqrt(x^6-9))+C

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integral of (2x^2)/(x^3sqrt(x^6-9))

\int \frac{2 x ^{2}}{x ^{3} x ^{6} - 9} d x

\frac{2}{9} arctan (\frac{1}{3} x^{6} - 9) + C

Solution steps

\int \frac{2 x ^{2}}{x ^{3} x ^{6} - 9} 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 ^{2}}{x ^{3} x ^{6} - 9} d x

Simplify

\frac{x ^{2}}{x ^{3} x ^{6} - 9} : \frac{1}{x x ^{6} - 9}

= 2 \cdot \int \frac{1}{x x ^{6} - 9} d x

Apply u-substitution

= 2 \cdot \int \frac{1}{6 u u - 9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u u - 9} d u

Apply u-substitution

= 2 \cdot \frac{1}{6} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \frac{1}{3} arctan (\frac{x ^{6} - 9}{3})

Simplify

2 \cdot \frac{1}{6} \cdot 2 \cdot \frac{1}{3} arctan (\frac{x ^{6} - 9}{3}) : \frac{2}{9} arctan (\frac{1}{3} x^{6} - 9)

= \frac{2}{9} arctan (\frac{1}{3} x^{6} - 9)

Add a constant to the solution

= \frac{2}{9} arctan (\frac{1}{3} x^{6} - 9) + C

What is the integral of (2x^2)/(x^3sqrt(x^6-9)) ?
The integral of (2x^2)/(x^3sqrt(x^6-9)) is 2/9 arctan(1/3 sqrt(x^6-9))+C