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

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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