What is the integral of (x^2)/(9x^2+4) ?

The integral of (x^2)/(9x^2+4) is 1/27 (3x-2arctan(3/2 x))+C

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integral of (x^2)/(9x^2+4)

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

\frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x)) + C

Solution steps

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

9 x^{2} + 4 = 4 (\frac{9 x ^{2}}{4} + 1)

= \int \frac{x ^{2}}{4 ( \frac{9 x ^{2}}{4} + 1 )} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{x ^{2}}{\frac{9 x ^{2}}{4} + 1} d x

Apply Integral Substitution

= \frac{1}{4} \cdot \int \frac{8 u ^{2}}{27 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \frac{8}{27} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= \frac{1}{4} \cdot \frac{8}{27} \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} \cdot \frac{8}{27} (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= \frac{1}{4} \cdot \frac{8}{27} (- arctan (u) + u)

Substitute back

u = \frac{3}{2} x

= \frac{1}{4} \cdot \frac{8}{27} (- arctan (\frac{3}{2} x) + \frac{3}{2} x)

Simplify

\frac{1}{4} \cdot \frac{8}{27} (- arctan (\frac{3}{2} x) + \frac{3}{2} x) : \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x))

= \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x))

Add a constant to the solution

= \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x)) + C

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What is the integral of (x^2)/(9x^2+4) ?
The integral of (x^2)/(9x^2+4) is 1/27 (3x-2arctan(3/2 x))+C