如何解析求解∫√(x/(ax³+b))及定积分∫₀¹√(x/(ax³+b))dx？

Let's break down how to get that integral result you found on WolframAlpha—step by step, so it makes sense why the log term pops up. We'll start with the indefinite integral, then move to the definite one from 0 to 1.

Step 1: Simplify the Integrand First #

First, rewrite the messy square root to make substitution easier:
$$\sqrt{\frac{x}{ax^3 + b}} = \frac{\sqrt{x}}{\sqrt{ax^3 + b}} = \frac{x^{1/2}}{\sqrt{b + ax^3}}$$

Step 2: Choose a Smart Substitution #

The key here is picking a substitution that turns the $ax^3$ term into something we can handle with standard integrals. Let's set:
$$t = x^{3/2}$$
Take the derivative with respect to $x$:
$$\frac{dt}{dx} = \frac{3}{2}x^{1/2} \implies x^{1/2}dx = \frac{2}{3}dt$$

Now substitute this into the original integral—notice how the $x^{1/2}dx$ term exactly matches what we have:
$$\int \frac{x^{1/2}}{\sqrt{b + ax^3}} dx = \int \frac{2}{3} \cdot \frac{dt}{\sqrt{b + a t^2}}$$

Step 3: Use a Standard Integral Formula #

We can rewrite the denominator to fit the form of a well-known integral. Factor out $\sqrt{a}$ from the square root:
$$\sqrt{b + a t^2} = \sqrt{a} \cdot \sqrt{t^2 + \frac{b}{a}}$$

Substitute that back in, and the integral becomes:
$$\frac{2}{3\sqrt{a}} \int \frac{dt}{\sqrt{t^2 + \left(\sqrt{\frac{b}{a}}\right)^2}}$$

Now we use the standard integral result (for real $a>0, b>0$):
$$\int \frac{du}{\sqrt{u^2 + c^2}} = \ln\left(u + \sqrt{u^2 + c^2}\right) + C$$

Applying this to our integral, we get:
$$\frac{2}{3\sqrt{a}} \ln\left(t + \sqrt{t^2 + \frac{b}{a}}\right) + C$$

Step 4: Substitute Back to $x$ #

Replace $t$ with $x^{3/2}$:
$$\frac{2}{3\sqrt{a}} \ln\left(x^{3/2} + \sqrt{x^3 + \frac{b}{a}}\right) + C$$

Let's clean up the square root term inside the log to match WolframAlpha's result. Multiply the numerator and denominator inside the square root by $a$:
$$\sqrt{x^3 + \frac{b}{a}} = \frac{\sqrt{ax^3 + b}}{\sqrt{a}}$$

Substitute that in:
$$\frac{2}{3\sqrt{a}} \ln\left(x^{3/2} + \frac{\sqrt{ax^3 + b}}{\sqrt{a}}\right) + C$$

Combine the terms inside the log over a common denominator:
$$\frac{2}{3\sqrt{a}} \ln\left(\frac{a x^{3/2} + \sqrt{ax^3 + b}}{\sqrt{a}}\right) + C$$

Using logarithm rules ($\ln(\frac{A}{B}) = \ln A - \ln B$), the $\ln\sqrt{a}$ term is a constant, so we can absorb it into the integration constant $C$:
$$\frac{2}{3\sqrt{a}} \ln\left(a x^{3/2} + \sqrt{ax^3 + b}\right) + C'$$

That's exactly the simplified form WolframAlpha gave! The first form they showed is just a redundant way to write the same thing—if you multiply out $\sqrt{\frac{x}{ax^3+b}} \cdot \sqrt{ax^3+b} / \sqrt{x}$, it cancels out to 1, so both expressions are equivalent.

Step 5: Compute the Definite Integral from 0 to 1 #

Now we evaluate the antiderivative at the bounds $x=1$ and $x=0$ (taking the limit as $x\to0^+$ since the antiderivative is defined there):

1. At $x=1$:
   $$\frac{2}{3\sqrt{a}} \ln\left(a \cdot 1^{3/2} + \sqrt{a \cdot 1^3 + b}\right) = \frac{2}{3\sqrt{a}} \ln\left(a + \sqrt{a + b}\right)$$

2. At $x\to0^+$:
   $$\lim_{x\to0^+} \frac{2}{3\sqrt{a}} \ln\left(a x^{3/2} + \sqrt{ax^3 + b}\right) = \frac{2}{3\sqrt{a}} \ln(\sqrt{b}) = \frac{1}{3\sqrt{a}} \ln b$$

Subtract the lower bound result from the upper bound result:
$$\int_0^1 \sqrt{\frac{x}{ax^3 + b}} dx = \frac{2}{3\sqrt{a}} \left[ \ln\left(a + \sqrt{a + b}\right) - \frac{1}{2}\ln b \right]$$

You can also rewrite this using logarithm properties if you prefer:
$$= \frac{2}{3\sqrt{a}} \ln\left( \frac{a + \sqrt{a + b}}{\sqrt{b}} \right)$$

Important Notes #

• This derivation assumes $a>0$ and $b>0$ to keep all square roots and logarithms real. If $a$ or $b$ are negative, you'd need to extend to complex numbers or adjust the substitution accordingly.

内容的提问来源于stack exchange，提问作者Study Astrophysics