What is the integral of 1/(x^2+10x-6) ?

The integral of 1/(x^2+10x-6) is -1/(2sqrt(31))(ln|(x+5)/(sqrt(31))+1|-ln|(x+5)/(sqrt(31))-1|)+C

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integral of 1/(x^2+10x-6)

\int \frac{1}{x ^{2} + 10 x - 6} d x

- \frac{1}{2 31} (ln \frac{x + 5}{31} + 1 - ln \frac{x + 5}{31} - 1) + C

Solution steps

\int \frac{1}{x ^{2} + 10 x - 6} d x

Complete the square

x^{2} + 10 x - 6 : (x + 5)^{2} - 31

= \int \frac{1}{( x + 5 ) ^{2} - 31} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 31} d u

= \int \frac{1}{u ^{2} - 31} d u

Apply Integral Substitution

= \int \frac{1}{31 ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{31} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{31} \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

= \frac{1}{31} (- \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}

= \frac{1}{31} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{31} - \frac{ln \frac{x + 5}{31} + 1}{2} - \frac{ln \frac{x + 5}{31} - 1}{2}

Simplify

\frac{1}{31} - \frac{ln \frac{x + 5}{31} + 1}{2} - \frac{ln \frac{x + 5}{31} - 1}{2} : - \frac{1}{2 31} (ln \frac{x + 5}{31} + 1 - ln \frac{x + 5}{31} - 1)

= - \frac{1}{2 31} (ln \frac{x + 5}{31} + 1 - ln \frac{x + 5}{31} - 1)

Add a constant to the solution

= - \frac{1}{2 31} (ln \frac{x + 5}{31} + 1 - ln \frac{x + 5}{31} - 1) + C

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What is the integral of 1/(x^2+10x-6) ?
The integral of 1/(x^2+10x-6) is -1/(2sqrt(31))(ln|(x+5)/(sqrt(31))+1|-ln|(x+5)/(sqrt(31))-1|)+C