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

The integral of 3/(sqrt(1-3x-x^2)) is 3arcsin(1/(sqrt(13))(2x+3))+C

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

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

3 arcsin (\frac{1}{13} (2 x + 3)) + C

Solution steps

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

Complete the square

1 - 3 x - x^{2} : - (x + \frac{3}{2})^{2} + \frac{13}{4}

= 3 \cdot \int \frac{1}{- ( x + \frac{3}{2} ) ^{2} + \frac{13}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 13} d u

= 3 \cdot \int \frac{2}{- 4 u ^{2} + 13} d u

Take the constant out:

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

= 3 \cdot 2 \cdot \int \frac{1}{- 4 u ^{2} + 13} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

= 3 \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 3 \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{13} (x + \frac{3}{2}))

Simplify

3 \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{13} (x + \frac{3}{2})) : 3 arcsin (\frac{1}{13} (2 x + 3))

= 3 arcsin (\frac{1}{13} (2 x + 3))

Add a constant to the solution

= 3 arcsin (\frac{1}{13} (2 x + 3)) + C

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What is the integral of 3/(sqrt(1-3x-x^2)) ?
The integral of 3/(sqrt(1-3x-x^2)) is 3arcsin(1/(sqrt(13))(2x+3))+C