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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

x \to - 4 lim (x^{2} cos (\frac{5}{x}))

\int \frac{3 tan ( β ) - 4 cos ^{2} ( β )}{cos ( β )} d β

\frac{2 x d y}{d x} = 1 - y^{2}

a re a y = x^{2} - 3, y = 6

a re a 4 cos (7 x), 4 - 4 cos (7 x)

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