What is the integral of 1/(sqrt(-3x^2+3x+7)) ?

The integral of 1/(sqrt(-3x^2+3x+7)) is 1/(sqrt(3))arcsin((2sqrt(3))/(sqrt(31))x-sqrt(3/31))+C

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integral of 1/(sqrt(-3x^2+3x+7))

\int \frac{1}{- 3 x ^{2} + 3 x + 7} d x

\frac{1}{3} arcsin (\frac{2 3}{31} x - \frac{3}{31}) + C

Solution steps

\int \frac{1}{- 3 x ^{2} + 3 x + 7} d x

Complete the square

- 3 x^{2} + 3 x + 7 : - 3 (x - \frac{1}{2})^{2} + \frac{31}{4}

= \int \frac{1}{- 3 ( x - \frac{1}{2} ) ^{2} + \frac{31}{4}} d x

Apply u-substitution

: \int \frac{2}{- 12 u ^{2} + 31} d u

= \int \frac{2}{- 12 u ^{2} + 31} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{- 12 u ^{2} + 31} d u

Apply Trigonometric Substitution

: \int \frac{1}{2 3} d v

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

= 2 \cdot \frac{1}{2 3} \cdot 1 \cdot arcsin (\frac{2 3}{31} (x - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2 3} \cdot 1 \cdot arcsin (\frac{2 3}{31} (x - \frac{1}{2})) : \frac{1}{3} arcsin (\frac{2 3}{31} x - \frac{3}{31})

= \frac{1}{3} arcsin (\frac{2 3}{31} x - \frac{3}{31})

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{2 3}{31} x - \frac{3}{31}) + C

x \to 2 lim (\frac{1}{2 ( x - 2 )})

\int \frac{1}{5 - 4 x - x ^{2}} d x

\int cos (4 x) d x

\int \frac{( x + 1 )}{x ^{2} - 4 x + 12} d x

f (x) = cos (3 x)

What is the integral of 1/(sqrt(-3x^2+3x+7)) ?
The integral of 1/(sqrt(-3x^2+3x+7)) is 1/(sqrt(3))arcsin((2sqrt(3))/(sqrt(31))x-sqrt(3/31))+C