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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

f (x) = x - 2, at x = 3

cos (2 x) + sin (x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 0

\int_{0}^{L} x d x

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