What is the integral of (3x^2+2x+1)/(x^2(x+1)) ?

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

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integral of (3x^2+2x+1)/(x^2(x+1))

\int \frac{3 x ^{2} + 2 x + 1}{x ^{2} ( x + 1 )} d x

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

Solution steps

\int \frac{3 x ^{2} + 2 x + 1}{x ^{2} ( x + 1 )} d x

Expand

\frac{3 x ^{2} + 2 x + 1}{x ^{2} ( x + 1 )} : \frac{3 x ^{2}}{x ^{2} ( x + 1 )} + \frac{2 x}{x ^{2} ( x + 1 )} + \frac{1}{x ^{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{3 x ^{2}}{x ^{2} ( x + 1 )} d x + \int \frac{2 x}{x ^{2} ( x + 1 )} d x + \int \frac{1}{x ^{2} ( x + 1 )} d x

\int \frac{3 x ^{2}}{x ^{2} ( x + 1 )} d x = 3 ln ∣ x + 1 ∣

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

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

= 3 ln ∣ x + 1 ∣ + 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) - ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣

Simplify

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

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

Add a constant to the solution

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

\int - sin^{2} (x) + cos^{2} (x) d x

\int_{1}^{3} \frac{x ^{2} - 1}{x} d x

\int arctan (x - 1) d x

\int \frac{1}{( x ^{3} - 1 )} d x

\int \frac{2 x ^{2} - 1}{x ( x + 1 ) ^{3}} d x

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