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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x \to 0 + lim (x^{n})

\frac{\partial}{\partial x} (8 x e^{x^{2}} + y^{2})

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

(x + 2)^{3}, 2 x^{2} - 8 x + 8, 0

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