To verify that the solution to the integral
∫ d x ( x − 3 ) x + 1 = 1 2 ln ∣ x + 1 − 2 x + 1 + 2 ∣ + C \int \frac{dx}{(x-3)\sqrt{x+1}} = \frac{1}{2} \ln \left| \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right| + C ∫ ( x − 3 ) x + 1 d x = 2 1 ln x + 1 + 2 x + 1 − 2 + C
is correct, we can differentiate the result and see if we get back the original integrand. Let's differentiate the result step by step.
Given:
F ( x ) = 1 2 ln ∣ x + 1 − 2 x + 1 + 2 ∣ + C F(x) = \frac{1}{2} \ln \left| \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right| + C F ( x ) = 2 1 ln x + 1 + 2 x + 1 − 2 + C
We need to find d F d x \frac{dF}{dx} d x d F
F ′ ( x ) = 1 2 d d x ln ∣ x + 1 − 2 x + 1 + 2 ∣ F'(x) = \frac{1}{2} \frac{d}{dx} \ln \left| \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right| F ′ ( x ) = 2 1 d x d ln x + 1 + 2 x + 1 − 2
Using the chain rule for differentiation:
F ′ ( x ) = 1 2 ⋅ 1 x + 1 − 2 x + 1 + 2 ⋅ d d x ( x + 1 − 2 x + 1 + 2 ) F'(x) = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2}} \cdot \frac{d}{dx} \left( \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right) F ′ ( x ) = 2 1 ⋅ x + 1 + 2 x + 1 − 2 1 ⋅ d x d ( x + 1 + 2 x + 1 − 2 )
Simplify the first part:
F ′ ( x ) = 1 2 ⋅ x + 1 + 2 x + 1 − 2 ⋅ d d x ( x + 1 − 2 x + 1 + 2 ) F'(x) = \frac{1}{2} \cdot \frac{\sqrt{x+1} + 2}{\sqrt{x+1} - 2} \cdot \frac{d}{dx} \left( \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right) F ′ ( x ) = 2 1 ⋅ x + 1 − 2 x + 1 + 2 ⋅ d x d ( x + 1 + 2 x + 1 − 2 )
Next, differentiate the quotient using the quotient rule ( u v ) ′ = u ′ v − u v ′ v 2 \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ u = x + 1 − 2 u = \sqrt{x+1} - 2 u = x + 1 − 2 v = x + 1 + 2 v = \sqrt{x+1} + 2 v = x + 1 + 2
u ′ = 1 2 x + 1 , v ′ = 1 2 x + 1 u' = \frac{1}{2\sqrt{x+1}}, \quad v' = \frac{1}{2\sqrt{x+1}} u ′ = 2 x + 1 1 , v ′ = 2 x + 1 1
Apply the quotient rule:
d d x ( x + 1 − 2 x + 1 + 2 ) = ( 1 2 x + 1 ) ( x + 1 + 2 ) − ( x + 1 − 2 ) ( 1 2 x + 1 ) ( x + 1 + 2 ) 2 \frac{d}{dx} \left( \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right) = \frac{(\frac{1}{2\sqrt{x+1}})(\sqrt{x+1} + 2) - (\sqrt{x+1} - 2)(\frac{1}{2\sqrt{x+1}})}{(\sqrt{x+1} + 2)^2} d x d ( x + 1 + 2 x + 1 − 2 ) = ( x + 1 + 2 ) 2 ( 2 x + 1 1 ) ( x + 1 + 2 ) − ( x + 1 − 2 ) ( 2 x + 1 1 )
Simplify the numerator:
= x + 1 + 2 2 x + 1 − x + 1 − 2 2 x + 1 ( x + 1 + 2 ) 2 = x + 1 + 2 − x + 1 + 2 2 x + 1 ( x + 1 + 2 ) 2 = \frac{\frac{\sqrt{x+1} + 2}{2\sqrt{x+1}} - \frac{\sqrt{x+1} - 2}{2\sqrt{x+1}}}{(\sqrt{x+1} + 2)^2} = \frac{\frac{\sqrt{x+1} + 2 - \sqrt{x+1} + 2}{2\sqrt{x+1}}}{(\sqrt{x+1} + 2)^2} = ( x + 1 + 2 ) 2 2 x + 1 x + 1 + 2 − 2 x + 1 x + 1 − 2 = ( x + 1 + 2 ) 2 2 x + 1 x + 1 + 2 − x + 1 + 2
= 4 2 x + 1 ( x + 1 + 2 ) 2 = 2 x + 1 ( x + 1 + 2 ) 2 = \frac{\frac{4}{2\sqrt{x+1}}}{(\sqrt{x+1} + 2)^2} = \frac{2}{\sqrt{x+1} (\sqrt{x+1} + 2)^2} = ( x + 1 + 2 ) 2 2 x + 1 4 = x + 1 ( x + 1 + 2 ) 2 2
Thus,
F ′ ( x ) = 1 2 ⋅ x + 1 + 2 x + 1 − 2 ⋅ 2 x + 1 ( x + 1 + 2 ) 2 F'(x) = \frac{1}{2} \cdot \frac{\sqrt{x+1} + 2}{\sqrt{x+1} - 2} \cdot \frac{2}{\sqrt{x+1} (\sqrt{x+1} + 2)^2} F ′ ( x ) = 2 1 ⋅ x + 1 − 2 x + 1 + 2 ⋅ x + 1 ( x + 1 + 2 ) 2 2
Cancel the common terms:
F ′ ( x ) = 1 2 ⋅ 1 x + 1 − 2 ⋅ 2 x + 1 ( x + 1 + 2 ) F'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{x+1} - 2} \cdot \frac{2}{\sqrt{x+1} (\sqrt{x+1} + 2)} F ′ ( x ) = 2 1 ⋅ x + 1 − 2 1 ⋅ x + 1 ( x + 1 + 2 ) 2
= 1 ( x + 1 − 2 ) ( x + 1 ) ( x + 1 + 2 ) = \frac{1}{(\sqrt{x+1} - 2)(\sqrt{x+1})(\sqrt{x+1} + 2)} = ( x + 1 − 2 ) ( x + 1 ) ( x + 1 + 2 ) 1
Simplify:
= 1 x + 1 ( x + 1 − 2 ) ( x + 1 + 2 ) = 1 ( x + 1 − 4 ) x + 1 = 1 ( x − 3 ) x + 1 = \frac{1}{\sqrt{x+1}(\sqrt{x+1} - 2)(\sqrt{x+1} + 2)} = \frac{1}{(x+1-4)\sqrt{x+1}} = \frac{1}{(x-3)\sqrt{x+1}} = x + 1 ( x + 1 − 2 ) ( x + 1 + 2 ) 1 = ( x + 1 − 4 ) x + 1 1 = ( x − 3 ) x + 1 1
Therefore, we have shown that:
F ′ ( x ) = d d x ( 1 2 ln ∣ x + 1 − 2 x + 1 + 2 ∣ + C ) = 1 ( x − 3 ) x + 1 F'(x) = \frac{d}{dx} \left( \frac{1}{2} \ln \left| \frac{\sqrt{x+1} - 2}{\sqrt{x+1} + 2} \right| + C \right) = \frac{1}{(x-3)\sqrt{x+1}} F ′ ( x ) = d x d ( 2 1 ln x + 1 + 2 x + 1 − 2 + C ) = ( x − 3 ) x + 1 1
This confirms that our original solution is correct.
If you have any questions or need further explanations, please let me know. Here are some other questions you might have:
Can you provide another example of verifying an integral solution?
How do you apply the chain rule in more complex differentiation problems?
Can you explain the quotient rule with more examples?
How do logarithmic functions impact differentiation?
Can you show another example of integral using substitution?
Tip: Always verify your integral solutions by differentiating your result and ensuring it matches the original integrand. This helps ensure accuracy and understanding.