What is the integral of (5x^2)/(x^2+7) ?

The integral of (5x^2)/(x^2+7) is -5sqrt(7)arctan(x/(sqrt(7)))+5x+C

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integral of (5x^2)/(x^2+7)

\int \frac{5 x ^{2}}{x ^{2} + 7} d x

- 57 arctan (\frac{x}{7}) + 5 x + C

Solution steps

\int \frac{5 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

= 5 \cdot \int \frac{x ^{2}}{x ^{2} + 7} d x

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

= 5 \cdot \int - \frac{7}{x ^{2} + 7} + 1 d x

Apply Integral Substitution

= 5 \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

= 57 \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

= 57 (- \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

= 57 (- arctan (u) + u)

Substitute back

u = \frac{x}{7}

= 57 (- arctan (\frac{x}{7}) + \frac{x}{7})

Simplify

57 (- arctan (\frac{x}{7}) + \frac{x}{7}) : - 57 arctan (\frac{x}{7}) + 5 x

= - 57 arctan (\frac{x}{7}) + 5 x

Add a constant to the solution

= - 57 arctan (\frac{x}{7}) + 5 x + C

\int sin^{3} (a x) d x

sin (x), sin (2 x), [0, \frac{π}{3}]

f (x) = \frac{1}{x} - 2 x

\frac{d}{d x} (\frac{3}{4} x^{2} - 5 x^{3})

\int \frac{1}{- 2 x} d x

What is the integral of (5x^2)/(x^2+7) ?
The integral of (5x^2)/(x^2+7) is -5sqrt(7)arctan(x/(sqrt(7)))+5x+C