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 e^{x^{3}} 3 x^{2} d x

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

\int 3 x + a d x

\int \frac{arcsin ^{4} ( x )}{1 - x ^{2}} d x

\int x (x + 2)^{9} 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