Question

Find : \(\displaystyle\int \dfrac{x+2}{\sqrt{9x-x^{2}}}\,dx\)

Hint:

Write x + 2 as -1/2 times (9 - 2x) plus 13/2, then use substitution for one part and complete the square for the other.

Answer 1

Step 1: Plan the split:
The expression under the root is \(9x-x^{2}\). Its derivative is \(9-2x\).
So we write the numerator as \(x+2 = A(9-2x)+B\).
Comparing the coefficients of \(x\) gives \(-2A=1\), so \(A=-\dfrac{1}{2}\).
Comparing the constants gives \(9A+B=2\), so \(B=2+\dfrac{9}{2}=\dfrac{13}{2}\).
Hence \(x+2 = -\dfrac{1}{2}(9-2x)+\dfrac{13}{2}\).

Step 2: Break the integral into two parts:
Let \(I\) be the required integral. Then
\[ I = -\frac{1}{2}\int \frac{9-2x}{\sqrt{9x-x^{2}}}\,dx + \frac{13}{2}\int \frac{dx}{\sqrt{9x-x^{2}}} = -\frac{1}{2}I_1 + \frac{13}{2}I_2 \]

Step 3: Find the first integral by substitution:
Put \(u = 9x-x^{2}\). Then \(du = (9-2x)\,dx\).
\[ I_1 = \int u^{-1/2}\,du = 2\sqrt{u} = 2\sqrt{9x-x^{2}} \]
So the first part is \(-\dfrac{1}{2}\cdot 2\sqrt{9x-x^{2}} = -\sqrt{9x-x^{2}}\).

Step 4: Find the second integral by completing the square:
\[ 9x-x^{2} = \left(\frac{9}{2}\right)^{2} - \left(x-\frac{9}{2}\right)^{2} \]
Use the standard result \(\displaystyle\int \frac{dt}{\sqrt{a^{2}-t^{2}}} = \sin^{-1}\frac{t}{a}\) with \(t = x-\dfrac{9}{2}\) and \(a=\dfrac{9}{2}\).
\[ I_2 = \sin^{-1}\left(\frac{x-\frac{9}{2}}{\frac{9}{2}}\right) = \sin^{-1}\left(\frac{2x-9}{9}\right) \]

Step 5: Join the parts:
\[ I = -\sqrt{9x-x^{2}} + \frac{13}{2}\sin^{-1}\left(\frac{2x-9}{9}\right) + C \]
Differentiating this result gives back \(\dfrac{x+2}{\sqrt{9x-x^{2}}}\), which confirms it.

Final Answer:
The integral equals \(-\sqrt{9x-x^{2}} + \dfrac{13}{2}\sin^{-1}\dfrac{2x-9}{9} + C\).\[ \boxed{-\sqrt{9x-x^{2}} + \frac{13}{2}\sin^{-1}\left(\frac{2x-9}{9}\right) + C} \]