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

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

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

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

\frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 1 ) ( x ^{2} + 4 )} : \frac{1}{5 ( x + 1 )} + \frac{- x + 1}{5 ( x ^{2} + 4 )}

= \int \frac{1}{5 ( x + 1 )} + \frac{- x + 1}{5 ( x ^{2} + 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 \frac{1}{5 ( x + 1 )} d x + \int \frac{- x + 1}{5 ( x ^{2} + 4 )} d x

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

\int \frac{- x + 1}{5 ( x ^{2} + 4 )} d x = \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}))

= \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})) + C

- 6 y^{^{''}} + 54 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

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

f (x) = (3 t - 1)^{4} (2 t + 1)^{- 3}

\int \frac{1}{( 4 - x ) ^{\frac{3}{2}}} d x

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