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)

integral

\frac{( 3 x ^{2} + 1 )}{x ^{3} + 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{4 x}{x ^{2} + 1}

\frac{x ^{2}}{x ^{2} + 21}

- 2 a^{7}

\frac{2}{( e ^{x} + e ^{- x} ) ^{2}}

\frac{1}{t ^{2}} \cdot \frac{e ^{- 1}}{t}

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