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

The integral of (x^2+x+1)/((x+2)(x^2+1)) is 3/5 ln|x+2|+1/5 (ln|x^2+1|+arctan(x))+C

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

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

\frac{3}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + x + 1}{( x + 2 ) ( x ^{2} + 1 )} : \frac{3}{5 ( x + 2 )} + \frac{2 x + 1}{5 ( x ^{2} + 1 )}

= \int \frac{3}{5 ( x + 2 )} + \frac{2 x + 1}{5 ( x ^{2} + 1 )} 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 \frac{3}{5 ( x + 2 )} d x + \int \frac{2 x + 1}{5 ( x ^{2} + 1 )} d x

\int \frac{3}{5 ( x + 2 )} d x = \frac{3}{5} ln ∣ x + 2 ∣

\int \frac{2 x + 1}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (ln x^{2} + 1 + arctan (x))

= \frac{3}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x))

Add a constant to the solution

= \frac{3}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x)) + C

x \to 7 lim (x^{2})

\frac{d}{d x} (cos^{3} (x^{2}))

x \to 2 lim ((2 x + 4)^{3})

x \to 0 + lim (x^{\frac{7}{x}})

\int sin (\frac{1}{6} x) d x

What is the integral of (x^2+x+1)/((x+2)(x^2+1)) ?
The integral of (x^2+x+1)/((x+2)(x^2+1)) is 3/5 ln|x+2|+1/5 (ln|x^2+1|+arctan(x))+C