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

The integral of (2+x)/(4x^2-4x-3) is -3/16 ln|2x+1|+7/16 ln|2x-3|+C

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

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

- \frac{3}{16} ln ∣ 2 x + 1 ∣ + \frac{7}{16} ln ∣ 2 x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 + x}{4 x ^{2} - 4 x - 3} : - \frac{3}{8 ( 2 x + 1 )} + \frac{7}{8 ( 2 x - 3 )}

= \int - \frac{3}{8 ( 2 x + 1 )} + \frac{7}{8 ( 2 x - 3 )} d x

Apply the Sum Rule:

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

= - \int \frac{3}{8 ( 2 x + 1 )} d x + \int \frac{7}{8 ( 2 x - 3 )} d x

\int \frac{3}{8 ( 2 x + 1 )} d x = \frac{3}{16} ln ∣ 2 x + 1 ∣

\int \frac{7}{8 ( 2 x - 3 )} d x = \frac{7}{16} ln ∣ 2 x - 3 ∣

= - \frac{3}{16} ln ∣ 2 x + 1 ∣ + \frac{7}{16} ln ∣ 2 x - 3 ∣

Add a constant to the solution

= - \frac{3}{16} ln ∣ 2 x + 1 ∣ + \frac{7}{16} ln ∣ 2 x - 3 ∣ + C

f (x) = ln (x - 1)

\int (12 x^{2} - 8 x + 2) d x

\frac{d y}{d x} = \frac{( x - 1 )}{y}

\frac{d}{d x} (7 arcsin (x^{2}))

y = 1 x, (0, 4)

What is the integral of (2+x)/(4x^2-4x-3) ?
The integral of (2+x)/(4x^2-4x-3) is -3/16 ln|2x+1|+7/16 ln|2x-3|+C