What is the integral of (3x^2+x+4)/(x^3+x) ?

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

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integral of (3x^2+x+4)/(x^3+x)

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n}

f (x) = 270

y = x^{3} - 4 x^{2} + 3 x, y = x^{2} - x

\frac{d y}{d x} = x + 6 y

\frac{\partial}{\partial x} (cos (x + 5 y))

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