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

The integral of (x^2)/((9+x^2)^2) is 1/6 arctan(x/3)-x/(2(9+x^2))+C

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

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

\frac{1}{6} arctan (\frac{x}{3}) - \frac{x}{2 ( 9 + x ^{2} )} + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Take the partial fraction of

\frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} : \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}}

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

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

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

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

= \frac{1}{3} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x}{3}

= \frac{1}{3} (arctan (\frac{x}{3}) - \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}}))

Simplify

\frac{1}{3} (arctan (\frac{x}{3}) - \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}})) : \frac{1}{6} arctan (\frac{x}{3}) - \frac{x}{2 ( 9 + x ^{2} )}

= \frac{1}{6} arctan (\frac{x}{3}) - \frac{x}{2 ( 9 + x ^{2} )}

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x}{3}) - \frac{x}{2 ( 9 + x ^{2} )} + C

\frac{\partial}{\partial x} (\frac{25 x y}{x - y})

tan (\frac{π x}{2})

y^{^{''}} + 4 y = - 5 csc (2 t)

\int \frac{1}{3 - x ^{2}} d x

\frac{\partial}{\partial y} (y + 2)

What is the integral of (x^2)/((9+x^2)^2) ?
The integral of (x^2)/((9+x^2)^2) is 1/6 arctan(x/3)-x/(2(9+x^2))+C