What is the integral of (1-x)/(4x^2-4x-3) ?

The integral of (1-x)/(4x^2-4x-3) is 1/16 (-2ln|4x^2-4x-3|-ln|x+1/2 |+ln|x-3/2 |)+C

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integral of (1-x)/(4x^2-4x-3)

\int \frac{1 - x}{4 x ^{2} - 4 x - 3} d x

\frac{1}{16} (- 2 ln 4 x^{2} - 4 x - 3 - ln x + \frac{1}{2} + ln x - \frac{3}{2}) + C

Solution steps

\int \frac{1 - x}{4 x ^{2} - 4 x - 3} d x

Complete the square

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

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{- 2 u + 1}{4 u ^{2} - 4} d u

Expand

\frac{- 2 u + 1}{4 u ^{2} - 4} : - \frac{2 u}{4 u ^{2} - 4} + \frac{1}{4 u ^{2} - 4}

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}{2} (- \int \frac{2 u}{4 u ^{2} - 4} d u + \int \frac{1}{4 u ^{2} - 4} d u)

\int \frac{2 u}{4 u ^{2} - 4} d u = \frac{1}{4} ln 4 u^{2} - 4

\int \frac{1}{4 u ^{2} - 4} d u = - \frac{1}{8} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)

= \frac{1}{2} (- \frac{1}{4} ln 4 u^{2} - 4 - \frac{1}{8} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣))

Substitute back

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

= \frac{1}{2} (- \frac{1}{4} ln 4 (x - \frac{1}{2})^{2} - 4 - \frac{1}{8} (ln x - \frac{1}{2} + 1 - ln x - \frac{1}{2} - 1))

Simplify

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

= \frac{1}{16} (- 2 ln 4 x^{2} - 4 x - 3 - ln x + \frac{1}{2} + ln x - \frac{3}{2})

Add a constant to the solution

= \frac{1}{16} (- 2 ln 4 x^{2} - 4 x - 3 - ln x + \frac{1}{2} + ln x - \frac{3}{2}) + C

What is the integral of (1-x)/(4x^2-4x-3) ?
The integral of (1-x)/(4x^2-4x-3) is 1/16 (-2ln|4x^2-4x-3|-ln|x+1/2 |+ln|x-3/2 |)+C