What is the integral of (6(x^2+4))/(x(x^2+6)) ?

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

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integral of (6(x^2+4))/(x(x^2+6))

\int \frac{6 ( x ^{2} + 4 )}{x ( x ^{2} + 6 )} d x

ln x^{2} + 6 + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{6 ( x ^{2} + 4 )}{x ( x ^{2} + 6 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2} + 4}{x ( x ^{2} + 6 )} d x

Expand

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

Apply the Sum Rule:

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

= 6 (\int \frac{x ^{2}}{x ( x ^{2} + 6 )} d x + \int \frac{4}{x ( x ^{2} + 6 )} d x)

\int \frac{x ^{2}}{x ( x ^{2} + 6 )} d x = \frac{1}{2} ln x^{2} + 6

\int \frac{4}{x ( x ^{2} + 6 )} d x = 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6)

= 6 (\frac{1}{2} ln x^{2} + 6 + 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6))

Simplify

6 (\frac{1}{2} ln x^{2} + 6 + 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6)) : ln x^{2} + 6 + 4 ln ∣ x ∣

= ln x^{2} + 6 + 4 ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 6 + 4 ln ∣ x ∣ + C

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\int \frac{3}{e ^{2 a} - 4} d a

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