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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{3 ( u ^{2} + 9 )} 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} + 9} 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{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

y = x^{2}, y = 2 x

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

\int x^{13} e^{x^{14}} d x

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

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