What is the integral of (x^2)/((x^2+16)^2) ?

The integral of (x^2)/((x^2+16)^2) is 1/4 (1/2 arctan(x/4)-(2x)/(16+x^2))+C

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integral of (x^2)/((x^2+16)^2)

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{2}} d x

\frac{1}{4} (\frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}}) + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{u ^{2}}{4 ( 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}{4} \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}{4} \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}{4} (\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}{4} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x}{4}

= \frac{1}{4} (arctan (\frac{x}{4}) - \frac{1}{2} (arctan (\frac{x}{4}) + \frac{\frac{x}{4}}{1 + ( \frac{x}{4} ) ^{2}}))

Simplify

\frac{1}{4} (arctan (\frac{x}{4}) - \frac{1}{2} (arctan (\frac{x}{4}) + \frac{\frac{x}{4}}{1 + ( \frac{x}{4} ) ^{2}})) : \frac{1}{4} (\frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}})

= \frac{1}{4} (\frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}})

Add a constant to the solution

= \frac{1}{4} (\frac{1}{2} arctan (\frac{x}{4}) - \frac{2 x}{16 + x ^{2}}) + C

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What is the integral of (x^2)/((x^2+16)^2) ?
The integral of (x^2)/((x^2+16)^2) is 1/4 (1/2 arctan(x/4)-(2x)/(16+x^2))+C