What is the integral of (6x^2+8)/(2x^3+8x) ?

The integral of (6x^2+8)/(2x^3+8x) is ln|x^2+4|+ln|x|+C

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integral of (6x^2+8)/(2x^3+8x)

\int \frac{6 x ^{2} + 8}{2 x ^{3} + 8 x} d x

ln x^{2} + 4 + ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} + 8}{2 x ^{3} + 8 x} d x

Simplify

\frac{6 x ^{2} + 8}{2 x ^{3} + 8 x} : \frac{3 x ^{2} + 4}{x ( x ^{2} + 4 )}

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

Expand

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

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

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

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

= \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Simplify

\frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : ln x^{2} + 4 + ln ∣ x ∣

= ln x^{2} + 4 + ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 4 + ln ∣ x ∣ + C

\int - 2 x e^{3 x} d x

\int \frac{1}{x ^{2} - 2 x - 4} d x

\int \frac{1}{x 25 x ^{2} - 5} d x

\int (3 x^{\frac{1}{2}} - x^{- 2} + 2) d x

What is the integral of (6x^2+8)/(2x^3+8x) ?
The integral of (6x^2+8)/(2x^3+8x) is ln|x^2+4|+ln|x|+C