What is the integral of x/(36x^4+9) ?

The integral of x/(36x^4+9) is 1/36 arctan(2x^2)+C

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integral of x/(36x^4+9)

\int \frac{x}{36 x ^{4} + 9} d x

\frac{1}{36} arctan (2 x^{2}) + C

Solution steps

\int \frac{x}{36 x ^{4} + 9} d x

Apply u-substitution

: \int \frac{1}{2 ( 36 u ^{2} + 9 )} d u

= \int \frac{1}{2 ( 36 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}{2} \cdot \int \frac{1}{36 u ^{2} + 9} d u

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{1}{9 v ^{2} + 9} = \frac{1}{9} \frac{1}{v ^{2} + 1}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{9} \cdot \frac{1}{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}{2} \cdot \frac{1}{2} \cdot \frac{1}{9} \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}{2} \cdot \frac{1}{2} \cdot \frac{1}{9} arctan (v)

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{9} arctan (\frac{6}{3} x^{2}) : \frac{1}{36} arctan (2 x^{2})

= \frac{1}{36} arctan (2 x^{2})

Add a constant to the solution

= \frac{1}{36} arctan (2 x^{2}) + C

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What is the integral of x/(36x^4+9) ?
The integral of x/(36x^4+9) is 1/36 arctan(2x^2)+C