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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

= x - ln ∣ x + 1 ∣ - \frac{1}{x + 1}

Add a constant to the solution

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

\int \frac{x ^{2}}{( x - 1 ) ^{100}} d x

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sec^{3} (x)

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