What is the integral of+(x^2)/((225+x^2)^2) ?

The integral of+(x^2)/((225+x^2)^2) is 1/30 arctan(x/(15))-x/(2(225+x^2))+C

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integral of+(x^2)/((225+x^2)^2)

\int + \frac{x ^{2}}{( 225 + x ^{2} ) ^{2}} d x

\frac{1}{30} arctan (\frac{x}{15}) - \frac{x}{2 ( 225 + x ^{2} )} + C

Solution steps

\int \frac{x ^{2}}{( 225 + x ^{2} ) ^{2}} d x

Apply Integral Substitution

= \int \frac{u ^{2}}{15 ( 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}{15} \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}{15} \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}{15} (\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}{15} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x}{15}

= \frac{1}{15} (arctan (\frac{x}{15}) - \frac{1}{2} (arctan (\frac{x}{15}) + \frac{\frac{x}{15}}{1 + ( \frac{x}{15} ) ^{2}}))

Simplify

\frac{1}{15} (arctan (\frac{x}{15}) - \frac{1}{2} (arctan (\frac{x}{15}) + \frac{\frac{x}{15}}{1 + ( \frac{x}{15} ) ^{2}})) : \frac{1}{30} arctan (\frac{x}{15}) - \frac{x}{2 ( 225 + x ^{2} )}

= \frac{1}{30} arctan (\frac{x}{15}) - \frac{x}{2 ( 225 + x ^{2} )}

Add a constant to the solution

= \frac{1}{30} arctan (\frac{x}{15}) - \frac{x}{2 ( 225 + x ^{2} )} + C

x \to 0 lim (\frac{( 3 ^{x} - 4 ^{x} )}{x})

e^{9 x}

\frac{1}{3} sin (3 x)

12 x^{3} - 3 y^{2} x^{4} + 1 y^{^{'}} = 0, y (0) = 2

\frac{cos ( x )}{x ^{9}}

What is the integral of+(x^2)/((225+x^2)^2) ?
The integral of+(x^2)/((225+x^2)^2) is 1/30 arctan(x/(15))-x/(2(225+x^2))+C