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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} (x + 1)^{2} + C

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