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

The integral of (x^2+1)/((x+2)^2) is x+2-4ln|x+2|-5/(x+2)+C

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

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

x + 2 - 4 ln ∣ x + 2 ∣ - \frac{5}{x + 2} + C

Solution steps

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

Expand

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

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

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

= x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} - \frac{1}{x + 2}

Simplify

x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2} - \frac{1}{x + 2} : x + 2 - 4 ln ∣ x + 2 ∣ - \frac{5}{x + 2}

= x + 2 - 4 ln ∣ x + 2 ∣ - \frac{5}{x + 2}

Add a constant to the solution

= x + 2 - 4 ln ∣ x + 2 ∣ - \frac{5}{x + 2} + C

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What is the integral of (x^2+1)/((x+2)^2) ?
The integral of (x^2+1)/((x+2)^2) is x+2-4ln|x+2|-5/(x+2)+C