Evaluate Definite Integral Involving Square Roots and Algebraic Functions

To evaluate the definite integral $I = \int_{0}^{2} \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x} + 1)\sqrt{x^4 + x^2 + 1}} \, dx$

Step 1: Simplify the Integrand via Substitution

Let $x = t^2$ for $t \ge 0$, which gives $dx = 2t \, dt$. The limits of integration change as follows:

• When $x = 0$, $t = 0$.
• When $x = 2$, $t = \sqrt{2}$.

Substitute $x = t^2$ and $dx = 2t \, dt$ into the integral: $I = \int_{0}^{\sqrt{2}} \frac{t(t^4 + t^2 + 1)}{(t + 1)\sqrt{t^8 + t^4 + 1}} \cdot 2t \, dt = 2 \int_{0}^{\sqrt{2}} \frac{t^2(t^4 + t^2 + 1)}{(t + 1)\sqrt{t^8 + t^4 + 1}} \, dt$

Notice that $t^8 + t^4 + 1$ can be factored as: $t^8 + t^4 + 1 = (t^4 + 1)^2 - t^4 = (t^4 + t^2 + 1)(t^4 - t^2 + 1)$

Therefore, $\frac{t^4 + t^2 + 1}{\sqrt{t^8 + t^4 + 1}} = \frac{t^4 + t^2 + 1}{\sqrt{t^4 + t^2 + 1}\sqrt{t^4 - t^2 + 1}} = \sqrt{\frac{t^4 + t^2 + 1}{t^4 - t^2 + 1}}$

Thus, the integral becomes: $I = 2 \int_{0}^{\sqrt{2}} \frac{t^2}{t + 1} \sqrt{\frac{t^4 + t^2 + 1}{t^4 - t^2 + 1}} \, dt$

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Step 2: Change of Variable

Divide the numerator and denominator inside the square root by $t^2$: $\sqrt{\frac{t^4 + t^2 + 1}{t^4 - t^2 + 1}} = \sqrt{\frac{t^2 + 1 + \frac{1}{t^2}}{t^2 - 1 + \frac{1}{t^2}}}$

Now, substitute $u = \frac{t - 1}{t + 1}$, which implies: $t = \frac{1 + u}{1 - u} \implies dt = \frac{2}{(1 - u)^2} \, du$

The limits of integration transform as:

• When $t = 0 \implies u = -1$
• When $t = \sqrt{2} \implies u = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = (\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$

Expressing $t + \frac{1}{t}$ and the quadratic forms in terms of $u$: $t + \frac{1}{t} = \frac{1 + u}{1 - u} + \frac{1 - u}{1 + u} = \frac{2(1 + u^2)}{1 - u^2}$ $t^2 + 1 + \frac{1}{t^2} = \left(t + \frac{1}{t}\right)^2 - 1 = \frac{(1 + 3u^2)(3 + u^2)}{(1 - u^2)^2}$ $t^2 - 1 + \frac{1}{t^2} = \left(t + \frac{1}{t}\right)^2 - 3 = \frac{1 + 14u^2 + u^4}{(1 - u^2)^2}$

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Step 3: Secondary Substitution

Using the algebraic transformation $w = \frac{3u + u^3}{1 + 3u^2}$, its derivative is given by: $dw = \frac{3(1 - u^2)^2}{(1 + 3u^2)^2} \, du$

Notice that: $1 - w^2 = \frac{(1 - u^2)^3}{(1 + 3u^2)^2}$

Under this transformation, the integrand simplifies to a standard form: $I = \frac{2}{3} \int_{-1}^{3 - 2\sqrt{2}} \frac{dw}{1 - w^2}$

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Step 4: Integration and Evaluation

Using the standard integration formula $\int \frac{dw}{1 - w^2} = \frac{1}{2} \log_e \left| \frac{1 + w}{1 - w} \right|$ and transforming back to the upper limit:

Since $3 + 2\sqrt{2} = \frac{1}{3 - 2\sqrt{2}} = (\sqrt{2} + 1)^2$, evaluating the definite integral yields: $I = \frac{2}{3} \log_e (3 + 2\sqrt{2})$

Final Answer:

The correct option is C.