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

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

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

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

- \frac{2}{5} (ln (\frac{4}{5} ∣ x ∣) - ln \frac{4}{5} x - 2) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

2 x^{2} - 5 x : 2 (x - \frac{5}{4})^{2} - \frac{25}{8}

= 2 \cdot \int \frac{1}{2 ( x - \frac{5}{4} ) ^{2} - \frac{25}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} - 25} d u

= 2 \cdot \int \frac{8}{16 u ^{2} - 25} d u

Take the constant out:

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

= 2 \cdot 8 \cdot \int \frac{1}{16 u ^{2} - 25} d u

Apply Integral Substitution

= 2 \cdot 8 \cdot \int \frac{1}{20 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 2 \cdot 8 \cdot \frac{1}{20} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 2 \cdot 8 \cdot \frac{1}{20} \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

= 2 \cdot 8 \cdot \frac{1}{20} (- \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}

= 2 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( x - \frac{5}{4} ) - 1}{2}))

Simplify

2 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( x - \frac{5}{4} ) - 1}{2})) : - \frac{2}{5} (ln (\frac{4}{5} ∣ x ∣) - ln \frac{4}{5} x - 2)

= - \frac{2}{5} (ln (\frac{4}{5} ∣ x ∣) - ln \frac{4}{5} x - 2)

Add a constant to the solution

= - \frac{2}{5} (ln (\frac{4}{5} ∣ x ∣) - ln \frac{4}{5} x - 2) + C

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