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

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

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

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

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

Solution steps

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

9 - (2 x + 1)^{2} = 2 - x^{2} - x + 2

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

Take the constant out:

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

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

Complete the square

- x^{2} - x + 2 : - (x + \frac{1}{2})^{2} + \frac{9}{4}

= \frac{1}{2} \cdot \int \frac{1}{- ( x + \frac{1}{2} ) ^{2} + \frac{9}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 9} d u

= \frac{1}{2} \cdot \int \frac{2}{- 4 u ^{2} + 9} d u

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{- 4 u ^{2} + 9} d u

Apply Trigonometric Substitution

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

= \frac{1}{2} \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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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