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

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

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

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

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

Solution steps

\int \frac{2 x + 3}{x ^{2} + 3 x + 4} 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^{2} + 3 x + 4

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

Add a constant to the solution

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

\int - 0.1 x d x

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

\int (x + 2) e^{4 x} d x

\int 2 + y e^{x y} d x

\int \frac{x ^{2} - 8 x + 15}{x ^{2} - 6 x + 5} d x

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