Step 1: Verify the Identity Using Differentiation

Given the integral identity:

$\int \frac{x^2}{\sqrt{x - 1}} \, dx = \frac{2}{15} \sqrt{x - 1} \left( 3x^2 + 4x + 8 \right) + C$

To verify this identity, we differentiate the right-hand side with respect to $x$ and check if it matches the integrand on the left-hand side.

Let $F(x) = \frac{2}{15} \sqrt{x - 1} \left( 3x^2 + 4x + 8 \right)$.

Differentiate $F(x)$ with respect to $x$:

$\frac{dF(x)}{dx} = \frac{2}{15} \left[ \frac{d}{dx} \left( \sqrt{x - 1} \right) \cdot \left( 3x^2 + 4x + 8 \right) + \sqrt{x - 1} \cdot \frac{d}{dx} \left( 3x^2 + 4x + 8 \right) \right]$

1. Differentiate $\sqrt{x - 1}$:

   $\frac{d}{dx} \left( \sqrt{x - 1} \right) = \frac{1}{2\sqrt{x - 1}} \cdot \frac{d}{dx}(x - 1) = \frac{1}{2\sqrt{x - 1}}$

2. Differentiate $3x^2 + 4x + 8$:

   $\frac{d}{dx} \left( 3x^2 + 4x + 8 \right) = 6x + 4$

Substitute these derivatives back into the equation for $\frac{dF(x)}{dx}$:

$\frac{dF(x)}{dx} = \frac{2}{15} \left[ \frac{1}{2\sqrt{x - 1}} \cdot \left( 3x^2 + 4x + 8 \right) + \sqrt{x - 1} \cdot (6x + 4) \right]$

Simplify this expression:

$\frac{dF(x)}{dx} = \frac{2}{15} \left[ \frac{3x^2 + 4x + 8}{2\sqrt{x - 1}} + \sqrt{x - 1} (6x + 4) \right]$

Next, let's combine the terms over a common denominator $2\sqrt{x - 1}$:

$\frac{dF(x)}{dx} = \frac{2}{15} \cdot \frac{(3x^2 + 4x + 8) + 2(x - 1)(6x + 4)}{2\sqrt{x - 1}}$

Expand and simplify the numerator:

$2(x - 1)(6x + 4) = 12x^2 + 8x - 12x - 8 = 12x^2 - 4$

So the expression becomes:

$\frac{dF(x)}{dx} = \frac{2}{15} \cdot \frac{3x^2 + 4x + 8 + 12x^2 - 4}{2\sqrt{x - 1}}$

Simplify further:

$\frac{dF(x)}{dx} = \frac{2}{15} \cdot \frac{15x^2 + 4x + 4}{2\sqrt{x - 1}} = \frac{2}{15} \cdot \frac{15(x^2)}{2\sqrt{x - 1}} = \frac{x^2}{\sqrt{x - 1}}$

Thus, we have verified that:

$\frac{dF(x)}{dx} = \frac{x^2}{\sqrt{x - 1}}$

which confirms the original identity.

Step 2: Identify $f(u)$ Using the Given Substitution

Given the substitution $u = x - 1$, we have:

$du = dx$

The integral becomes:

$\int \frac{x^2}{\sqrt{x - 1}} \, dx = \int \frac{(u + 1)^2}{\sqrt{u}} \, du$

Expanding $(u + 1)^2$:

$(u + 1)^2 = u^2 + 2u + 1$

So the integral now is:

$\int \frac{u^2 + 2u + 1}{\sqrt{u}} \, du = \int \left( u^{3/2} + 2u^{1/2} + u^{-1/2} \right) du$

Here, $f(u) = u^{3/2} + 2u^{1/2} + u^{-1/2}$.

Summary

• The identity is verified by differentiation.
• For the given substitution $u = x - 1$, the function $f(u)$ is identified as:

$f(u) = u^{3/2} + 2u^{1/2} + u^{-1/2}$

Would you like more details or have any questions?

Here are five related questions:

1. How do you apply the chain rule in differentiation?
2. What are the common techniques for verifying identities using differentiation?
3. How does substitution work in definite and indefinite integrals?
4. What are the steps to solve integrals involving square roots?
5. How can the binomial theorem be applied to expand expressions like $(u + 1)^2$?

Tip: When verifying an identity using differentiation, ensure every step is clearly justified to avoid errors.