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

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

\frac{d}{d x} (e^{- 4 x} sin (4 x))

\int \frac{tan ^{2} ( θ )}{sec ( θ )} d θ

- y + \frac{d y}{d x} ln (x) = 0

\frac{d}{d x} (arcsin (x^{2} + 1))

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