What is the integral of 1/(sqrt(1-6x-9x^2)) ?

The integral of 1/(sqrt(1-6x-9x^2)) is 1/3 arcsin(1/(sqrt(2))(3x+1))+C

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integral of 1/(sqrt(1-6x-9x^2))

\int \frac{1}{1 - 6 x - 9 x ^{2}} d x

\frac{1}{3} arcsin (\frac{1}{2} (3 x + 1)) + C

Solution steps

\int \frac{1}{1 - 6 x - 9 x ^{2}} d x

Complete the square

1 - 6 x - 9 x^{2} : - 9 (x + \frac{1}{3})^{2} + 2

= \int \frac{1}{- 9 ( x + \frac{1}{3} ) ^{2} + 2} d x

Apply u-substitution

: \int \frac{1}{- 9 u ^{2} + 2} d u

= \int \frac{1}{- 9 u ^{2} + 2} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} (x + \frac{1}{3}))

Simplify

\frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} (x + \frac{1}{3})) : \frac{1}{3} arcsin (\frac{1}{2} (3 x + 1))

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

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{1}{2} (3 x + 1)) + C

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What is the integral of 1/(sqrt(1-6x-9x^2)) ?
The integral of 1/(sqrt(1-6x-9x^2)) is 1/3 arcsin(1/(sqrt(2))(3x+1))+C