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

The integral of (x^2)/(4x^2+9) is 1/8 (2x-3arctan(2/3 x))+C

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

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

\frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x)) + C

Solution steps

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

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

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{9} \cdot \int \frac{27 u ^{2}}{8 ( 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}{9} \cdot \frac{27}{8} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

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

= \frac{1}{9} \cdot \frac{27}{8} \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}{9} \cdot \frac{27}{8} (- \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}{9} \cdot \frac{27}{8} (- arctan (u) + u)

Substitute back

u = \frac{2}{3} x

= \frac{1}{9} \cdot \frac{27}{8} (- arctan (\frac{2}{3} x) + \frac{2}{3} x)

Simplify

\frac{1}{9} \cdot \frac{27}{8} (- arctan (\frac{2}{3} x) + \frac{2}{3} x) : \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x))

= \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x))

Add a constant to the solution

= \frac{1}{8} (2 x - 3 arctan (\frac{2}{3} x)) + C

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\frac{d}{d x} (- x + 4)

What is the integral of (x^2)/(4x^2+9) ?
The integral of (x^2)/(4x^2+9) is 1/8 (2x-3arctan(2/3 x))+C