What is the integral of (6(x^2+5))/(x(x^2+6)) ?

The integral of (6(x^2+5))/(x(x^2+6)) is 1/2 ln|x^2+6|+5ln|x|+C

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integral of (6(x^2+5))/(x(x^2+6))

\int \frac{6 ( x ^{2} + 5 )}{x ( x ^{2} + 6 )} d x

\frac{1}{2} ln x^{2} + 6 + 5 ln ∣ x ∣ + C

Solution steps

\int \frac{6 ( x ^{2} + 5 )}{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} + 5}{x ( x ^{2} + 6 )} d x

Expand

\frac{x ^{2} + 5}{x ( x ^{2} + 6 )} : \frac{x ^{2}}{x ( x ^{2} + 6 )} + \frac{5}{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{5}{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{5}{x ( x ^{2} + 6 )} d x = 5 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6)

= 6 (\frac{1}{2} ln x^{2} + 6 + 5 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6))

Simplify

6 (\frac{1}{2} ln x^{2} + 6 + 5 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 6)) : \frac{1}{2} ln x^{2} + 6 + 5 ln ∣ x ∣

= \frac{1}{2} ln x^{2} + 6 + 5 ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 + 5 ln ∣ x ∣ + C

\frac{1}{3} x^{\frac{3}{2}}

\int_{- \frac{1}{5}}^{\frac{1}{5}} \frac{1}{1 - x ^{2}} d x

\frac{3 ( y + 1 ) ^{2}}{x ^{4}} d x - \frac{( 2 x ) ( y + 1 )}{x ^{4}} d y = 0

10 \frac{e ^{- 3 s}}{s}

x \to 4 - lim (x - 4)

What is the integral of (6(x^2+5))/(x(x^2+6)) ?
The integral of (6(x^2+5))/(x(x^2+6)) is 1/2 ln|x^2+6|+5ln|x|+C