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

The integral of 7/(2x^2-5x-3) is -ln| 2/7+4/7 x|+ln|-12/7+4/7 x|+C

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

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

- ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x + C

Solution steps

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

Take the constant out:

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

= 7 \cdot \int \frac{1}{2 x ^{2} - 5 x - 3} d x

Complete the square

2 x^{2} - 5 x - 3 : 2 (x - \frac{5}{4})^{2} - \frac{49}{8}

= 7 \cdot \int \frac{1}{2 ( x - \frac{5}{4} ) ^{2} - \frac{49}{8}} d x

Apply u-substitution

= 7 \cdot \int \frac{8}{16 u ^{2} - 49} d u

Take the constant out:

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

= 7 \cdot 8 \cdot \int \frac{1}{16 u ^{2} - 49} d u

Apply Integral Substitution

= 7 \cdot 8 \cdot \int \frac{1}{28 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 7 \cdot 8 \cdot \frac{1}{28} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 7 \cdot 8 \cdot \frac{1}{28} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 7 \cdot 8 \cdot \frac{1}{28} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x - \frac{5}{4} ) - 1}{2}))

Simplify

7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x - \frac{5}{4} ) - 1}{2})) : - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x

= - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x

Add a constant to the solution

= - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x + C

What is the integral of 7/(2x^2-5x-3) ?
The integral of 7/(2x^2-5x-3) is -ln| 2/7+4/7 x|+ln|-12/7+4/7 x|+C