What is the integral of (6x+4)/(3x^2+4x+5) ?

The integral of (6x+4)/(3x^2+4x+5) is ln|3x^2+4x+5|+C

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integral of (6x+4)/(3x^2+4x+5)

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 5} d x

ln 3 x^{2} + 4 x + 5 + C

Solution steps

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 5} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 3 x^{2} + 4 x + 5

= ln 3 x^{2} + 4 x + 5

Add a constant to the solution

= ln 3 x^{2} + 4 x + 5 + C

What is the integral of (6x+4)/(3x^2+4x+5) ?
The integral of (6x+4)/(3x^2+4x+5) is ln|3x^2+4x+5|+C