What is the integral of (x^2)/(9+x^6) ?

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

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integral of (x^2)/(9+x^6)

\int \frac{x ^{2}}{9 + x ^{6}} d x

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

Solution steps

\int \frac{x ^{2}}{9 + x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 ( 9 + u ^{2} )} d u

= \int \frac{1}{3 ( 9 + 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}{9 + u ^{2}} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} arctan (\frac{x ^{3}}{3})

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{3} + 6 x}{x})

\int cos^{2} (a x) d x

n = 7 \sum \infty \frac{1}{( n - 4 ) ^{4}}

\int \frac{( x - 1 )}{x ^{2} + 5 x + 6} d x

y^{^{''}} - \frac{5}{t} y^{^{'}} + \frac{25}{t ^{2}} y = 0

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