Write the numerator as A ⋅ ( derivative of the quantity under the root ) + B A\cdot(\text{derivative of the quantity under the root})+B A ⋅ ( derivative of the quantity under the root ) + B 2 ⋯ 2\sqrt{\cdots} 2 ⋯
(i) Let I = ∫ 2 x + 1 9 + 4 x − x 2 d x I=\displaystyle\int\frac{2x+1}{\sqrt{9+4x-x^2}}\,dx I = ∫ 9 + 4 x − x 2 2 x + 1 d x
Step 1. Write 2 x + 1 = A ( 4 − 2 x ) + B 2x+1=A(4-2x)+B 2 x + 1 = A ( 4 − 2 x ) + B − 2 A = 2 ⇒ A = − 1 -2A=2\Rightarrow A=-1 − 2 A = 2 ⇒ A = − 1 4 A + B = 1 ⇒ B = 5 4A+B=1\Rightarrow B=5 4 A + B = 1 ⇒ B = 5
Step 2. I = − ∫ 4 − 2 x 9 + 4 x − x 2 d x + 5 ∫ d x 9 + 4 x − x 2 = − 2 9 + 4 x − x 2 + 5 ∫ d x 9 + 4 x − x 2 I=-\displaystyle\int\frac{4-2x}{\sqrt{9+4x-x^2}}\,dx+5\int\frac{dx}{\sqrt{9+4x-x^2}}=-2\sqrt{9+4x-x^2}+5\int\frac{dx}{\sqrt{9+4x-x^2}} I = − ∫ 9 + 4 x − x 2 4 − 2 x d x + 5 ∫ 9 + 4 x − x 2 d x = − 2 9 + 4 x − x 2 + 5 ∫ 9 + 4 x − x 2 d x
Step 3. 9 + 4 x − x 2 = − [ ( x − 2 ) 2 − 4 − 9 ] = 13 − ( x − 2 ) 2 9+4x-x^2=-\big[(x-2)^2-4-9\big]=13-(x-2)^2 9 + 4 x − x 2 = − [ ( x − 2 ) 2 − 4 − 9 ] = 13 − ( x − 2 ) 2 ∫ d x 13 − ( x − 2 ) 2 = sin − 1 ( x − 2 13 ) + c \displaystyle\int\frac{dx}{\sqrt{13-(x-2)^2}}=\sin^{-1}\!\left(\dfrac{x-2}{\sqrt{13}}\right)+c ∫ 13 − ( x − 2 ) 2 d x = sin − 1 ( 13 x − 2 ) + c I = − 2 9 + 4 x − x 2 + 5 sin − 1 ( x − 2 13 ) + c I=-2\sqrt{9+4x-x^2}+5\sin^{-1}\!\left(\dfrac{x-2}{\sqrt{13}}\right)+c I = − 2 9 + 4 x − x 2 + 5 sin − 1 ( 13 x − 2 ) + c
(ii) Let I = ∫ x + 2 x 2 − 1 d x = ∫ x x 2 − 1 d x + 2 ∫ d x x 2 − 1 I=\displaystyle\int\frac{x+2}{\sqrt{x^2-1}}\,dx=\int\frac{x}{\sqrt{x^2-1}}\,dx+2\int\frac{dx}{\sqrt{x^2-1}} I = ∫ x 2 − 1 x + 2 d x = ∫ x 2 − 1 x d x + 2 ∫ x 2 − 1 d x
Step 1. ∫ x x 2 − 1 d x = x 2 − 1 \displaystyle\int\frac{x}{\sqrt{x^2-1}}\,dx=\sqrt{x^2-1} ∫ x 2 − 1 x d x = x 2 − 1 ∫ f ′ ( x ) [ f ( x ) ] − 1 / 2 d x \int f'(x)[f(x)]^{-1/2}dx ∫ f ′ ( x ) [ f ( x ) ] − 1/2 d x f = x 2 − 1 f=x^2-1 f = x 2 − 1
Step 2. 2 ∫ d x x 2 − 1 = 2 log ∣ x + x 2 − 1 ∣ + c 2\displaystyle\int\frac{dx}{\sqrt{x^2-1}}=2\log\left|x+\sqrt{x^2-1}\right|+c 2 ∫ x 2 − 1 d x = 2 log x + x 2 − 1 + c I = x 2 − 1 + 2 log ∣ x + x 2 − 1 ∣ + c I=\sqrt{x^2-1}+2\log\left|x+\sqrt{x^2-1}\right|+c I = x 2 − 1 + 2 log x + x 2 − 1 + c
(iii) Let I = ∫ 2 x + 3 x 2 + 4 x + 1 d x I=\displaystyle\int\frac{2x+3}{\sqrt{x^2+4x+1}}\,dx I = ∫ x 2 + 4 x + 1 2 x + 3 d x