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

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

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

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

ln x^{3} + x + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{3} + x

= ln x^{3} + x

Add a constant to the solution

= ln x^{3} + x + C

\frac{d}{d x} f (x) = \frac{2}{5} x^{5} - 4 x^{3} + 9 x^{2} - 6 x - 2

y = 2 x^{- \frac{3}{4}}

t \to 0 lim (\frac{t ^{2} + 9 - 3}{t ^{2}})

\int \frac{1}{2 x - 1} - \frac{1}{2 x - 1} d x

x y^{^{'}} = \frac{x ^{3} cos ( 2 x )}{y} + y

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