What is the integral of (4x^2)/(x^2+6) ?

The integral of (4x^2)/(x^2+6) is -4sqrt(6)arctan(x/(sqrt(6)))+4x+C

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integral of (4x^2)/(x^2+6)

\int \frac{4 x ^{2}}{x ^{2} + 6} d x

- 46 arctan (\frac{x}{6}) + 4 x + C

Solution steps

\int \frac{4 x ^{2}}{x ^{2} + 6} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{x ^{2} + 6} d x

\frac{x ^{2}}{x ^{2} + 6} = - \frac{6}{x ^{2} + 6} + 1

= 4 \cdot \int - \frac{6}{x ^{2} + 6} + 1 d x

Apply Integral Substitution

= 4 \cdot \int 6 (- \frac{1}{u ^{2} + 1} + 1) d u

Take the constant out:

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

= 46 \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

= 46 (- \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

= 46 (- arctan (u) + u)

Substitute back

u = \frac{x}{6}

= 46 (- arctan (\frac{x}{6}) + \frac{x}{6})

Simplify

46 (- arctan (\frac{x}{6}) + \frac{x}{6}) : - 46 arctan (\frac{x}{6}) + 4 x

= - 46 arctan (\frac{x}{6}) + 4 x

Add a constant to the solution

= - 46 arctan (\frac{x}{6}) + 4 x + C

y^{^{'}} - y = 3

\int \frac{12}{x ^{2} + 5 x + 6} d x

y^{^{''}} + 49 y = 2 tan (7 x)

\frac{d}{d x} (\frac{cos ( x )}{1 + sin ( x )})

\frac{d x}{d y} = \frac{y}{2}

What is the integral of (4x^2)/(x^2+6) ?
The integral of (4x^2)/(x^2+6) is -4sqrt(6)arctan(x/(sqrt(6)))+4x+C