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

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

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

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

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

Solution steps

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

\frac{x ^{2}}{x ^{2} + 9} = - \frac{9}{x ^{2} + 9} + 1

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

Apply Integral Substitution

= \int 3 (- \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

= 3 \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

= 3 (- \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

= 3 (- arctan (u) + u)

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

x \to - π - lim (csc (x))

x^{^{''}} + 2 x^{^{'}} + 2 = 0

n = 0 \sum \infty \frac{10}{3 ^{n}}

\frac{\partial}{\partial x} ((x + y)^{p} - b x - (1 - b) y)

\int z z^{2} - 1 d z

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