What is the integral of (5(x^2+4))/(x(x^2+5)) ?

The integral of (5(x^2+4))/(x(x^2+5)) is 1/2 ln|x^2+5|+4ln|x|+C

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integral of (5(x^2+4))/(x(x^2+5))

\int \frac{5 ( x ^{2} + 4 )}{x ( x ^{2} + 5 )} d x

\frac{1}{2} ln x^{2} + 5 + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{5 ( x ^{2} + 4 )}{x ( x ^{2} + 5 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2} + 4}{x ( x ^{2} + 5 )} d x

Expand

\frac{x ^{2} + 4}{x ( x ^{2} + 5 )} : \frac{x ^{2}}{x ( x ^{2} + 5 )} + \frac{4}{x ( x ^{2} + 5 )}

Apply the Sum Rule:

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

= 5 (\int \frac{x ^{2}}{x ( x ^{2} + 5 )} d x + \int \frac{4}{x ( x ^{2} + 5 )} d x)

\int \frac{x ^{2}}{x ( x ^{2} + 5 )} d x = \frac{1}{2} ln x^{2} + 5

\int \frac{4}{x ( x ^{2} + 5 )} d x = 4 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5)

= 5 (\frac{1}{2} ln x^{2} + 5 + 4 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5))

Simplify

5 (\frac{1}{2} ln x^{2} + 5 + 4 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln x^{2} + 5)) : \frac{1}{2} ln x^{2} + 5 + 4 ln ∣ x ∣

= \frac{1}{2} ln x^{2} + 5 + 4 ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 5 + 4 ln ∣ x ∣ + C

x \to 0 lim (+ (sin (x)))

f (x) = x + 8, at x = 1

x \to 1 lim (2^{x^{3} - x})

x \to 7 + lim (\frac{( x - 8 )}{x - 7})

\int \frac{1}{1 + 25 x ^{2}} d x

What is the integral of (5(x^2+4))/(x(x^2+5)) ?
The integral of (5(x^2+4))/(x(x^2+5)) is 1/2 ln|x^2+5|+4ln|x|+C