What is the integral of 1/(sqrt(5-7x-3x^2)) ?

The integral of 1/(sqrt(5-7x-3x^2)) is 1/(sqrt(3))arcsin((6x+7)/(sqrt(109)))+C

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integral of 1/(sqrt(5-7x-3x^2))

\int \frac{1}{5 - 7 x - 3 x ^{2}} d x

\frac{1}{3} arcsin (\frac{6 x + 7}{109}) + C

Solution steps

\int \frac{1}{5 - 7 x - 3 x ^{2}} d x

Complete the square

5 - 7 x - 3 x^{2} : - 3 (x + \frac{7}{6})^{2} + \frac{109}{12}

= \int \frac{1}{- 3 ( x + \frac{7}{6} ) ^{2} + \frac{109}{12}} d x

Apply u-substitution

= \int \frac{2 3}{- 36 u ^{2} + 109} d u

Take the constant out:

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

= 23 \cdot \int \frac{1}{- 36 u ^{2} + 109} d u

Apply Trigonometric Substitution

= 23 \cdot \int \frac{1}{6} d v

Take the constant out:

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

= 23 \frac{1}{6} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 23 \frac{1}{6} \cdot 1 \cdot v

Substitute back

= 23 \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{109} (x + \frac{7}{6}))

Simplify

23 \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{109} (x + \frac{7}{6})) : \frac{1}{3} arcsin (\frac{6 x + 7}{109})

= \frac{1}{3} arcsin (\frac{6 x + 7}{109})

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{6 x + 7}{109}) + C

What is the integral of 1/(sqrt(5-7x-3x^2)) ?
The integral of 1/(sqrt(5-7x-3x^2)) is 1/(sqrt(3))arcsin((6x+7)/(sqrt(109)))+C