What is the integral of (x^3)/((4+x^2)^2) ?

The integral of (x^3)/((4+x^2)^2) is 1/2 (ln|4+x^2|+4/(4+x^2))+C

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integral of (x^3)/((4+x^2)^2)

\int \frac{x ^{3}}{( 4 + x ^{2} ) ^{2}} d x

\frac{1}{2} (ln 4 + x^{2} + \frac{4}{4 + x ^{2}}) + C

Solution steps

\int \frac{x ^{3}}{( 4 + x ^{2} ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 4}{2 u ^{2}} d u

= \int \frac{u - 4}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u - 4}{u ^{2}} d u

Expand

\frac{u - 4}{u ^{2}} : \frac{1}{u} - \frac{4}{u ^{2}}

= \frac{1}{2} \cdot \int \frac{1}{u} - \frac{4}{u ^{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}{2} (\int \frac{1}{u} d u - \int \frac{4}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{4}{u ^{2}} d u = - \frac{4}{u}

= \frac{1}{2} (ln ∣ u ∣ - (- \frac{4}{u}))

Substitute back

u = 4 + x^{2}

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} (ln 4 + x^{2} + \frac{4}{4 + x ^{2}}) + C

\int \frac{1}{x ^{2} - 9 x + 18} d x

\int \frac{25 - x ^{2}}{x ^{2}} d x

y = \frac{x ^{10} - 5 x ^{6} - 12 x}{x ^{6}}

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}, y (1) = 1

\frac{\partial}{\partial x} (r^{2} + s^{2})

What is the integral of (x^3)/((4+x^2)^2) ?
The integral of (x^3)/((4+x^2)^2) is 1/2 (ln|4+x^2|+4/(4+x^2))+C