What is the integral of 1/(1-6x-3x^2) ?

The integral of 1/(1-6x-3x^2) is 1/(4sqrt(3))(ln|(sqrt(3))/2 (x+1)+1|-ln|(sqrt(3))/2 (x+1)-1|)+C

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integral of 1/(1-6x-3x^2)

\int \frac{1}{1 - 6 x - 3 x ^{2}} d x

\frac{1}{4 3} (ln \frac{3}{2} (x + 1) + 1 - ln \frac{3}{2} (x + 1) - 1) + C

Solution steps

\int \frac{1}{1 - 6 x - 3 x ^{2}} d x

Factor

1 - 6 x - 3 x^{2} : - (- 1 + 6 x + 3 x^{2})

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

Take the constant out:

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

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

Complete the square

- 1 + 6 x + 3 x^{2} : 3 (x + 1)^{2} - 4

= - \int \frac{1}{3 ( x + 1 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} - 4} d u

= - \int \frac{1}{3 u ^{2} - 4} d u

Apply Integral Substitution

= - \int \frac{1}{2 3 ( 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}{2 3} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

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

Substitute back

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

Simplify

- \frac{1}{2 3} - \frac{ln \frac{3}{2} ( x + 1 ) + 1}{2} - \frac{ln \frac{3}{2} ( x + 1 ) - 1}{2} : \frac{1}{4 3} (ln \frac{3}{2} (x + 1) + 1 - ln \frac{3}{2} (x + 1) - 1)

= \frac{1}{4 3} (ln \frac{3}{2} (x + 1) + 1 - ln \frac{3}{2} (x + 1) - 1)

Add a constant to the solution

= \frac{1}{4 3} (ln \frac{3}{2} (x + 1) + 1 - ln \frac{3}{2} (x + 1) - 1) + C

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What is the integral of 1/(1-6x-3x^2) ?
The integral of 1/(1-6x-3x^2) is 1/(4sqrt(3))(ln|(sqrt(3))/2 (x+1)+1|-ln|(sqrt(3))/2 (x+1)-1|)+C