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

The integral of (3x^3+8x)/((x^2+2)^2) is 3/2 ln|x^2+2|-1/(x^2+2)+C

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

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

\frac{3}{2} ln x^{2} + 2 - \frac{1}{x ^{2} + 2} + C

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{8 x}{( x ^{2} + 2 ) ^{2}} d x = - \frac{4}{x ^{2} + 2}

= \frac{3}{2} ln x^{2} + 2 + \frac{3}{x ^{2} + 2} - \frac{4}{x ^{2} + 2}

Simplify

= \frac{3}{2} ln x^{2} + 2 - \frac{1}{x ^{2} + 2}

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 2 - \frac{1}{x ^{2} + 2} + C

\frac{\partial}{\partial x} (p (q, x) q)

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

\int \frac{1}{x} x^{'} d x

\frac{d}{d x} (sin (\frac{x}{1 + y}))

\frac{d}{d x} (5 π cos (π x))

What is the integral of (3x^3+8x)/((x^2+2)^2) ?
The integral of (3x^3+8x)/((x^2+2)^2) is 3/2 ln|x^2+2|-1/(x^2+2)+C