What is the integral of (3x^2)/(x^2+7) ?

The integral of (3x^2)/(x^2+7) is -3sqrt(7)arctan(x/(sqrt(7)))+3x+C

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integral of (3x^2)/(x^2+7)

\int \frac{3 x ^{2}}{x ^{2} + 7} d x

- 37 arctan (\frac{x}{7}) + 3 x + C

Solution steps

\int \frac{3 x ^{2}}{x ^{2} + 7} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{x ^{2} + 7} d x

\frac{x ^{2}}{x ^{2} + 7} = - \frac{7}{x ^{2} + 7} + 1

= 3 \cdot \int - \frac{7}{x ^{2} + 7} + 1 d x

Apply Integral Substitution

= 3 \cdot \int 7 (- \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

= 37 \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

= 37 (- \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

= 37 (- arctan (u) + u)

Substitute back

u = \frac{x}{7}

= 37 (- arctan (\frac{x}{7}) + \frac{x}{7})

Simplify

37 (- arctan (\frac{x}{7}) + \frac{x}{7}) : - 37 arctan (\frac{x}{7}) + 3 x

= - 37 arctan (\frac{x}{7}) + 3 x

Add a constant to the solution

= - 37 arctan (\frac{x}{7}) + 3 x + C

f (x) = (\frac{3}{x})

(4 - x)^{^{'}}

\int_{0}^{9} 9 x - x^{2} d x

x \to 5 lim (3 x)

h (x) = \frac{3}{x ^{3} + 1}

What is the integral of (3x^2)/(x^2+7) ?
The integral of (3x^2)/(x^2+7) is -3sqrt(7)arctan(x/(sqrt(7)))+3x+C