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

The integral of (4x+2)/(x^2+x+5) is 2ln|4x^2+4x+20|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 ln 4 x^{2} + 4 x + 20 + C

Solution steps

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

Complete the square

x^{2} + x + 5 : (x + \frac{1}{2})^{2} + \frac{19}{4}

= \int \frac{4 x + 2}{( x + \frac{1}{2} ) ^{2} + \frac{19}{4}} d x

Apply u-substitution

: \int \frac{16 u}{4 u ^{2} + 19} d u

= \int \frac{16 u}{4 u ^{2} + 19} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int \frac{u}{4 u ^{2} + 19} d u

Apply u-substitution

: \int \frac{1}{8 v} d v

= 16 \cdot \int \frac{1}{8 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 16 \cdot \frac{1}{8} ln ∣ v ∣

Substitute back

= 16 \cdot \frac{1}{8} ln 4 (x + \frac{1}{2})^{2} + 19

Simplify

16 \cdot \frac{1}{8} ln 4 (x + \frac{1}{2})^{2} + 19 : 2 ln 4 x^{2} + 4 x + 20

= 2 ln 4 x^{2} + 4 x + 20

Add a constant to the solution

= 2 ln 4 x^{2} + 4 x + 20 + C

5 x - x^{2} (1.4)

\frac{\partial}{\partial x} (x^{2} - 4 y^{2})

x \to + 1 lim (x^{3} - 1)

\frac{d ^{2} y}{d x ^{2}} - b y = 0

100 x - \frac{d y}{d x} = x y

What is the integral of (4x+2)/(x^2+x+5) ?
The integral of (4x+2)/(x^2+x+5) is 2ln|4x^2+4x+20|+C