What is the integral of ((x^2+1)/(x^3+3x)) ?

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

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integral of ((x^2+1)/(x^3+3x))

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

\frac{1}{3} ln x^{3} + 3 x + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 3 x

= \frac{1}{3} ln x^{3} + 3 x

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 3 x + C

y = \frac{4 x ^{2} + 8 x + 4}{x}

x \to 5 lim (5 - x)

k = 0 \sum \infty \frac{1}{2} \cdot (- 1)^{k}

\frac{\partial}{\partial x} (\frac{1}{1 + x ^{2}})

x \to 1 lim (1^{\frac{1}{1 - x}})

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