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

\frac{d}{d x} (x^{- 11})

\int \frac{10}{( 5 - 4 x - x ^{2} ) ^{\frac{5}{2}}} d x

y^{^{''}} + y = 2

y^{^{''}} + y^{^{'}} - 12 y = 0

\frac{\partial ^{2}}{\partial u ( v ) \partial v} (u (v) (v)^{2} e^{- v})

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