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

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

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

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

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