What is the integral of (4x+2)/(x^2+x+1) ?

The integral of (4x+2)/(x^2+x+1) is 2ln|4x^2+4x+4|+C

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integral of (4x+2)/(x^2+x+1)

\int \frac{4 x + 2}{x ^{2} + x + 1} d x

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

Solution steps

\int \frac{4 x + 2}{x ^{2} + x + 1} d x

Complete the square

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{4 x + 2}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Apply u-substitution

: \int \frac{16 u}{4 u ^{2} + 3} d u

= \int \frac{16 u}{4 u ^{2} + 3} 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} + 3} 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} + 3

Simplify

16 \cdot \frac{1}{8} ln 4 (x + \frac{1}{2})^{2} + 3 : 2 ln 4 x^{2} + 4 x + 4

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

Add a constant to the solution

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

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What is the integral of (4x+2)/(x^2+x+1) ?
The integral of (4x+2)/(x^2+x+1) is 2ln|4x^2+4x+4|+C