What is the integral of ((x+1)^2)/((x^2+2x+4)) ?

The integral of ((x+1)^2)/((x^2+2x+4)) is x-sqrt(3)arctan(1/(sqrt(3))(x+1))+C

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integral of ((x+1)^2)/((x^2+2x+4))

\int \frac{( x + 1 ) ^{2}}{( x ^{2} + 2 x + 4 )} d x

x - 3 arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{( x + 1 ) ^{2}}{x ^{2} + 2 x + 4} d x

Long division

\frac{( x + 1 ) ^{2}}{x ^{2} + 2 x + 4} : 1 - \frac{3}{x ^{2} + 2 x + 4}

= \int 1 - \frac{3}{x ^{2} + 2 x + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int \frac{3}{x ^{2} + 2 x + 4} d x

\int 1 d x = x

\int \frac{3}{x ^{2} + 2 x + 4} d x = 3 arctan (\frac{1}{3} (x + 1))

= x - 3 arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= x - 3 arctan (\frac{1}{3} (x + 1)) + C

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What is the integral of ((x+1)^2)/((x^2+2x+4)) ?
The integral of ((x+1)^2)/((x^2+2x+4)) is x-sqrt(3)arctan(1/(sqrt(3))(x+1))+C