What is the integral of (x^2)/(5x^2+12) ?

The integral of (x^2)/(5x^2+12) is -(2sqrt(3))/(5sqrt(5))arctan((sqrt(5))/(2sqrt(3))x)+x/5+C

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integral of (x^2)/(5x^2+12)

\int \frac{x ^{2}}{5 x ^{2} + 12} d x

- \frac{2 3}{5 5} arctan (\frac{5}{2 3} x) + \frac{x}{5} + C

Solution steps

\int \frac{x ^{2}}{5 x ^{2} + 12} d x

5 x^{2} + 12 = 12 (\frac{5 x ^{2}}{12} + 1)

= \int \frac{x ^{2}}{12 ( \frac{5 x ^{2}}{12} + 1 )} d x

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{x ^{2}}{\frac{5 x ^{2}}{12} + 1} d x

Apply Integral Substitution

= \frac{1}{12} \cdot \int \frac{24 3 u ^{2}}{5 5 ( 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}{12} \cdot \frac{24 3}{5 5} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

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

= \frac{1}{12} \cdot \frac{24 3}{5 5} \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}{12} \cdot \frac{24 3}{5 5} (- \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}{12} \cdot \frac{24 3}{5 5} (- arctan (u) + u)

Substitute back

u = \frac{5}{2 3} x

= \frac{1}{12} \cdot \frac{24 3}{5 5} (- arctan (\frac{5}{2 3} x) + \frac{5}{2 3} x)

Simplify

\frac{1}{12} \cdot \frac{24 3}{5 5} (- arctan (\frac{5}{2 3} x) + \frac{5}{2 3} x) : - \frac{2 3}{5 5} arctan (\frac{5}{2 3} x) + \frac{x}{5}

= - \frac{2 3}{5 5} arctan (\frac{5}{2 3} x) + \frac{x}{5}

Add a constant to the solution

= - \frac{2 3}{5 5} arctan (\frac{5}{2 3} x) + \frac{x}{5} + C

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What is the integral of (x^2)/(5x^2+12) ?
The integral of (x^2)/(5x^2+12) is -(2sqrt(3))/(5sqrt(5))arctan((sqrt(5))/(2sqrt(3))x)+x/5+C