What is the integral of sqrt(9/((2+x)(3-x))) ?

The integral of sqrt(9/((2+x)(3-x))) is 6arcsin(1/(sqrt(5))sqrt(2+x))+C

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integral of sqrt(9/((2+x)(3-x)))

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

6 arcsin (\frac{1}{5} 2 + x) + C

Solution steps

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

Simplify

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

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

Apply u-substitution

= 3 \cdot \int \frac{1}{u ( - u + 5 )} d u

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

u (- u + 5) = u (- u + 5)

= 3 \cdot \int \frac{1}{u - u + 5} d u

Apply u-substitution

= 3 \cdot \int \frac{2}{- v ^{2} + 5} d v

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}{- v ^{2} + 5} d v

Apply Trigonometric Substitution

= 3 \cdot 2 \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 3 \cdot 2 \cdot 1 \cdot w

Substitute back

= 3 \cdot 2 \cdot 1 \cdot arcsin (\frac{1}{5} 2 + x)

Simplify

= 6 arcsin (\frac{1}{5} 2 + x)

Add a constant to the solution

= 6 arcsin (\frac{1}{5} 2 + x) + C

What is the integral of sqrt(9/((2+x)(3-x))) ?
The integral of sqrt(9/((2+x)(3-x))) is 6arcsin(1/(sqrt(5))sqrt(2+x))+C