What is the integral of (4(x^2+1))/(x(x^2+4)) ?

The integral of (4(x^2+1))/(x(x^2+4)) is 3/2 ln|x^2+4|+ln|x|+C

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integral of (4(x^2+1))/(x(x^2+4))

\int \frac{4 ( x ^{2} + 1 )}{x ( x ^{2} + 4 )} d x

\frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣ + C

Solution steps

\int \frac{4 ( x ^{2} + 1 )}{x ( x ^{2} + 4 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2} + 1}{x ( x ^{2} + 4 )} d x

Expand

\frac{x ^{2} + 1}{x ( x ^{2} + 4 )} : \frac{x ^{2}}{x ( x ^{2} + 4 )} + \frac{1}{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

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

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

\int \frac{1}{x ( x ^{2} + 4 )} d x = \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4

= 4 (\frac{1}{2} ln x^{2} + 4 + \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

4 (\frac{1}{2} ln x^{2} + 4 + \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣

= \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣

Add a constant to the solution

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

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What is the integral of (4(x^2+1))/(x(x^2+4)) ?
The integral of (4(x^2+1))/(x(x^2+4)) is 3/2 ln|x^2+4|+ln|x|+C