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

4 x - 6 y x^{2} + 1 (\frac{d y}{d x}) = 0

n = 1 \sum \infty \frac{1}{n ^{0.08}}

\frac{d}{d x} (a x e^{3 x})

3 x^{5}

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

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