What is the integral of 1/(2x^2+5x-1) ?

The integral of 1/(2x^2+5x-1) is -1/(sqrt(33))(ln|(4x+5)/(sqrt(33))+1|-ln|(4x+5)/(sqrt(33))-1|)+C

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integral of 1/(2x^2+5x-1)

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

- \frac{1}{33} (ln \frac{4 x + 5}{33} + 1 - ln \frac{4 x + 5}{33} - 1) + C

Solution steps

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

Complete the square

2 x^{2} + 5 x - 1 : 2 (x + \frac{5}{4})^{2} - \frac{33}{8}

= \int \frac{1}{2 ( x + \frac{5}{4} ) ^{2} - \frac{33}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} - 33} d u

= \int \frac{8}{16 u ^{2} - 33} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 u ^{2} - 33} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{4 33 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4 33} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 8 \cdot \frac{1}{4 33} \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

= 8 \cdot \frac{1}{4 33} (- \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}

= 8 \cdot \frac{1}{4 33} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 \cdot \frac{1}{4 33} - \frac{ln \frac{4}{33} ( x + \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{33} ( x + \frac{5}{4} ) - 1}{2}

Simplify

8 \cdot \frac{1}{4 33} - \frac{ln \frac{4}{33} ( x + \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{33} ( x + \frac{5}{4} ) - 1}{2} : - \frac{1}{33} (ln \frac{4 x + 5}{33} + 1 - ln \frac{4 x + 5}{33} - 1)

= - \frac{1}{33} (ln \frac{4 x + 5}{33} + 1 - ln \frac{4 x + 5}{33} - 1)

Add a constant to the solution

= - \frac{1}{33} (ln \frac{4 x + 5}{33} + 1 - ln \frac{4 x + 5}{33} - 1) + C

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What is the integral of 1/(2x^2+5x-1) ?
The integral of 1/(2x^2+5x-1) is -1/(sqrt(33))(ln|(4x+5)/(sqrt(33))+1|-ln|(4x+5)/(sqrt(33))-1|)+C