What is the integral of (6x)/(3x^2+1) ?

The integral of (6x)/(3x^2+1) is ln|3x^2+1|+C

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integral of (6x)/(3x^2+1)

\int \frac{6 x}{3 x ^{2} + 1} d x

ln 3 x^{2} + 1 + C

Solution steps

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

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

Simplify

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

= ln 3 x^{2} + 1

Add a constant to the solution

= ln 3 x^{2} + 1 + C

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What is the integral of (6x)/(3x^2+1) ?
The integral of (6x)/(3x^2+1) is ln|3x^2+1|+C