What is the integral of (-7x-7)/(2x^2+5x-3) ?

The integral of (-7x-7)/(2x^2+5x-3) is

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integral of (-7x-7)/(2x^2+5x-3)

\int \frac{- 7 x - 7}{2 x ^{2} + 5 x - 3} d x

- \frac{7}{4} ln 16 x^{2} + 40 x - 24 - \frac{5}{4} (ln \frac{4}{7} x + \frac{12}{7} - ln \frac{4}{7} x - \frac{2}{7}) - ln ∣ 2 x - 1 ∣ + ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{- 7 x - 7}{2 x ^{2} + 5 x - 3} d x

Expand

\frac{- 7 x - 7}{2 x ^{2} + 5 x - 3} : - \frac{7 x}{2 x ^{2} + 5 x - 3} - \frac{7}{2 x ^{2} + 5 x - 3}

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{7 x}{2 x ^{2} + 5 x - 3} d x - \int \frac{7}{2 x ^{2} + 5 x - 3} d x

\int \frac{7 x}{2 x ^{2} + 5 x - 3} d x = 14 (\frac{1}{8} ln 16 x^{2} + 40 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{4}{7} x - ln - \frac{2}{7} + \frac{4}{7} x))

\int \frac{7}{2 x ^{2} + 5 x - 3} d x = ln ∣ 2 x - 1 ∣ - ln ∣ x + 3 ∣

= - 14 (\frac{1}{8} ln 16 x^{2} + 40 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{4}{7} x - ln - \frac{2}{7} + \frac{4}{7} x)) - (ln ∣ 2 x - 1 ∣ - ln ∣ x + 3 ∣)

Simplify

- 14 (\frac{1}{8} ln 16 x^{2} + 40 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{4}{7} x - ln - \frac{2}{7} + \frac{4}{7} x)) - (ln ∣ 2 x - 1 ∣ - ln ∣ x + 3 ∣) : - \frac{7}{4} ln 16 x^{2} + 40 x - 24 - \frac{5}{4} (ln \frac{4}{7} x + \frac{12}{7} - ln \frac{4}{7} x - \frac{2}{7}) - ln ∣ 2 x - 1 ∣ + ln ∣ x + 3 ∣

= - \frac{7}{4} ln 16 x^{2} + 40 x - 24 - \frac{5}{4} (ln \frac{4}{7} x + \frac{12}{7} - ln \frac{4}{7} x - \frac{2}{7}) - ln ∣ 2 x - 1 ∣ + ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{7}{4} ln 16 x^{2} + 40 x - 24 - \frac{5}{4} (ln \frac{4}{7} x + \frac{12}{7} - ln \frac{4}{7} x - \frac{2}{7}) - ln ∣ 2 x - 1 ∣ + ln ∣ x + 3 ∣ + C

What is the integral of (-7x-7)/(2x^2+5x-3) ?
The integral of (-7x-7)/(2x^2+5x-3) is