What is the integral of (8x^3)/((x^2+2)^2) ?

The integral of (8x^3)/((x^2+2)^2) is 4(ln|x^2+2|+2/(x^2+2))+C

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integral of (8x^3)/((x^2+2)^2)

\int \frac{8 x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

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

Solution steps

\int \frac{8 x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

Apply u-substitution

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

= 8 \cdot \int \frac{u - 2}{2 u ^{2}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{u - 2}{u ^{2}} d u

Expand

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} - \frac{2}{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

= 8 \cdot \frac{1}{2} (\int \frac{1}{u} d u - \int \frac{2}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 8 \cdot \frac{1}{2} (ln ∣ u ∣ - (- \frac{2}{u}))

Substitute back

u = x^{2} + 2

= 8 \cdot \frac{1}{2} (ln x^{2} + 2 - (- \frac{2}{x ^{2} + 2}))

Simplify

8 \cdot \frac{1}{2} (ln x^{2} + 2 - (- \frac{2}{x ^{2} + 2})) : 4 (ln x^{2} + 2 + \frac{2}{x ^{2} + 2})

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

Add a constant to the solution

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

\int (\frac{8}{x}) d x

\frac{d y}{d x} = \frac{1}{x ^{8}}

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

\frac{d}{d x} (x ln (y) + sin (3 x))

n = 1 \sum \infty (\frac{10}{17})^{n}

What is the integral of (8x^3)/((x^2+2)^2) ?
The integral of (8x^3)/((x^2+2)^2) is 4(ln|x^2+2|+2/(x^2+2))+C