关于使用分部积分法求解不定积分∫dx/(x²+9)²的技术问询（禁用三角代换）

Hey there, let's work through this integral step by step using integration by parts—no trig substitution required at all. First, let's denote the integral we need to find as:
$$I = \int \frac{dx}{(x^2 + 9)^2}$$

The key here is to use integration by parts to relate (I) to a simpler integral we know how to handle. Let's start by applying integration by parts to the basic integral ( \int \frac{dx}{x^2 + 9} ):

Recall the integration by parts formula:
$$\int u , dv = uv - \int v , du$$

Choose (u = \frac{1}{x^2 + 9}) and (dv = dx). Now calculate (du) and (v):

• (du = \frac{d}{dx}\left(\frac{1}{x^2 + 9}\right) dx = \frac{-2x}{(x^2 + 9)^2} dx)
• (v = x)

Plug these into the formula:
$$\int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} - \int x \cdot \left( \frac{-2x}{(x^2 + 9)^2} \right) dx$$

Simplify the right-hand side integral term:
$$\int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} + 2 \int \frac{x^2}{(x2 + 9)^2} dx$$

Now rewrite (x^2) as ((x^2 + 9) - 9) to split the integral into two manageable parts:
$$\int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} + 2 \int \frac{(x^2 + 9) - 9}{(x^2 + 9)^2} dx$$

Split the integral:
$$\int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} + 2 \int \frac{dx}{x^2 + 9} - 18 \int \frac{dx}{(x^2 + 9)^2}$$

Notice that the last term on the right is (18I) (since (I = \int \frac{dx}{(x^2 + 9)^2})). Now move all terms with ( \int \frac{dx}{x^2 + 9} ) to the left-hand side:
$$\int \frac{dx}{x^2 + 9} - 2 \int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} - 18I$$

Simplify the left-hand side:
$$- \int \frac{dx}{x^2 + 9} = \frac{x}{x^2 + 9} - 18I$$

Now solve for (I):
$$18I = \frac{x}{x^2 + 9} + \int \frac{dx}{x^2 + 9}$$
$$I = \frac{1}{18} \left( \frac{x}{x^2 + 9} + \int \frac{dx}{x^2 + 9} \right)$$

While the integral ( \int \frac{dx}{x^2 + 9} ) is typically found using trig substitution, we don't need to use trigonometry to derive (I) itself. If you want the full closed-form solution, you can use the standard result for that basic integral (which doesn't require us to do trig substitution here):
$$\int \frac{dx}{x^2 + 9} = \frac{1}{3} \arctan\left( \frac{x}{3} \right) + C$$

Substitute that back in to get the final result:
$$I = \frac{x}{18(x^2 + 9)} + \frac{1}{54} \arctan\left( \frac{x}{3} \right) + C$$

And that's it—we solved the integral entirely with integration by parts and algebraic manipulation, no trigonometry involved in the derivation process.

备注：内容来源于stack exchange，提问作者Nil