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

The integral of x/(4x^2+4x+1) is 1/8 (2ln|x+1/2 |+2/(2x+1))+C

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

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

\frac{1}{8} (2 ln x + \frac{1}{2} + \frac{2}{2 x + 1}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 1}{8 u ^{2}} d u

= \int \frac{2 u - 1}{8 u ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{2 u - 1}{u ^{2}} d u

Expand

\frac{2 u - 1}{u ^{2}} : \frac{2}{u} - \frac{1}{u ^{2}}

= \frac{1}{8} \cdot \int \frac{2}{u} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int \frac{2}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= \frac{1}{8} (2 ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = x + \frac{1}{2}

= \frac{1}{8} (2 ln x + \frac{1}{2} - (- \frac{1}{x + \frac{1}{2}}))

Simplify

\frac{1}{8} (2 ln x + \frac{1}{2} - (- \frac{1}{x + \frac{1}{2}})) : \frac{1}{8} (2 ln x + \frac{1}{2} + \frac{2}{2 x + 1})

= \frac{1}{8} (2 ln x + \frac{1}{2} + \frac{2}{2 x + 1})

Add a constant to the solution

= \frac{1}{8} (2 ln x + \frac{1}{2} + \frac{2}{2 x + 1}) + C

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What is the integral of x/(4x^2+4x+1) ?
The integral of x/(4x^2+4x+1) is 1/8 (2ln|x+1/2 |+2/(2x+1))+C