What is the integral of (6x)/(2+3x^2) ?

The integral of (6x)/(2+3x^2) is ln|2+3x^2|+C

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integral of (6x)/(2+3x^2)

\int \frac{6 x}{2 + 3 x ^{2}} d x

ln 2 + 3 x^{2} + C

Solution steps

\int \frac{6 x}{2 + 3 x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{2 + 3 x ^{2}} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 6 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 2 + 3 x^{2}

= 6 \cdot \frac{1}{6} ln 2 + 3 x^{2}

Simplify

6 \cdot \frac{1}{6} ln 2 + 3 x^{2} : ln 2 + 3 x^{2}

= ln 2 + 3 x^{2}

Add a constant to the solution

= ln 2 + 3 x^{2} + C

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What is the integral of (6x)/(2+3x^2) ?
The integral of (6x)/(2+3x^2) is ln|2+3x^2|+C