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

The integral of (2x^2)/(9+x^2) is 6(-arctan(x/3)+x/3)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Apply Integral Substitution

= 2 \cdot \int 3 (- \frac{1}{1 + u ^{2}} + 1) d u

Take the constant out:

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

= 2 \cdot 3 \cdot \int - \frac{1}{1 + u ^{2}} + 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

= 2 \cdot 3 (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

\int 1 d u = u

= 2 \cdot 3 (- arctan (u) + u)

Substitute back

u = \frac{x}{3}

= 2 \cdot 3 (- arctan (\frac{x}{3}) + \frac{x}{3})

Simplify

= 6 (- arctan (\frac{x}{3}) + \frac{x}{3})

Add a constant to the solution

= 6 (- arctan (\frac{x}{3}) + \frac{x}{3}) + C

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What is the integral of (2x^2)/(9+x^2) ?
The integral of (2x^2)/(9+x^2) is 6(-arctan(x/3)+x/3)+C