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

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

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

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

\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x ∣ - 2 ln (2 ∣ x ∣) - \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 (ln ∣ x ∣ - ln ∣ 2 x + 1 ∣) + 2 ln ∣ 2 x + 1 ∣ - 2 ln (2 ∣ x ∣) - \frac{1}{x} : \frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x ∣ - 2 ln (2 ∣ x ∣) - \frac{1}{x}

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

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x ∣ - 2 ln (2 ∣ x ∣) - \frac{1}{x} + C

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