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

The integral of 3/(x^2+2x-5) is -(sqrt(3))/(2sqrt(2))(ln|(x+1)/(sqrt(6))+1|-ln|(x+1)/(sqrt(6))-1|)+C

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

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

- \frac{3}{2 2} (ln \frac{x + 1}{6} + 1 - ln \frac{x + 1}{6} - 1) + C

Solution steps

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

Take the constant out:

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

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

Factor

x^{2} + 2 x - 5 : - (- x^{2} - 2 x + 5)

= 3 \cdot \int \frac{1}{- ( - x ^{2} - 2 x + 5 )} d x

Take the constant out:

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

= 3 (- \int \frac{1}{- x ^{2} - 2 x + 5} d x)

Complete the square

- x^{2} - 2 x + 5 : - (x + 1)^{2} + 6

= 3 (- \int \frac{1}{- ( x + 1 ) ^{2} + 6} d x)

Apply u-substitution

: \int \frac{1}{- u ^{2} + 6} d u

= 3 (- \int \frac{1}{- u ^{2} + 6} d u)

Apply Integral Substitution

= 3 (- \int \frac{1}{6 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 3 (- \frac{1}{6} \cdot \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}

= 3 (- \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 3 - \frac{1}{6} \frac{ln \frac{x + 1}{6} + 1}{2} - \frac{ln \frac{x + 1}{6} - 1}{2}

Simplify

3 - \frac{1}{6} \frac{ln \frac{x + 1}{6} + 1}{2} - \frac{ln \frac{x + 1}{6} - 1}{2} : - \frac{3}{2 2} (ln \frac{x + 1}{6} + 1 - ln \frac{x + 1}{6} - 1)

= - \frac{3}{2 2} (ln \frac{x + 1}{6} + 1 - ln \frac{x + 1}{6} - 1)

Add a constant to the solution

= - \frac{3}{2 2} (ln \frac{x + 1}{6} + 1 - ln \frac{x + 1}{6} - 1) + C

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What is the integral of 3/(x^2+2x-5) ?
The integral of 3/(x^2+2x-5) is -(sqrt(3))/(2sqrt(2))(ln|(x+1)/(sqrt(6))+1|-ln|(x+1)/(sqrt(6))-1|)+C