求 $\int \frac{x}{\sqrt{2x-1}}\,dx.$ 利用换元法 $t = \tan x$，求 $\int \frac{1}{4\cos^{2}x + 9\sin^{2}x}\,dx.$ 上图显示了曲线 $y = x^{2} + 3x$ 的一部分，并用矩形近似表示从 $x = 0$ 到 $x = 1$ 的曲线下方面积。曲线下方的面积可以用 $(n-1)$ 个宽度为 $\tfrac{1}{n}$ 的矩形的总面积 $A$ 来近似表示。已知 $\sum_{r=1}^{n-1} r^{2} = \tfrac{1}{6} n(n+1)(2n+1),$ 证明 $A = \frac{(n-1)(11n-1)}{6n^{2}}$。 简要解释如何从该表达式推导出 $\int_{0}^{1} x^{2} + 3x\,dx$ 的值，并由此在不进行积分计算的情况下求出该面积的准确值。

1. Rewrite the numerator in terms of the expression in the denominator so that it can be split into integrable power terms.

$\int \frac{x}{\sqrt{2x-1}} \, dx = \int \frac{\frac{1}{2}(2x-1) + \frac{1}{2}}{\sqrt{2x-1}} \, dx$
$= \int \left[ \frac{1}{2}(2x-1)^{\frac{1}{2}} + \frac{1}{2}(2x-1)^{-\frac{1}{2}} \right] \, dx$

2. Integrate each term using the standard formula $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}$.

$= \frac{1}{2} \left[ \frac{(2x-1)^{\frac{3}{2}}}{\frac{3}{2} \cdot 2} \right] + \frac{1}{2} \left[ \frac{(2x-1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 2} \right] + C$
$= \frac{1}{6}(2x-1)^{\frac{3}{2}} + \frac{1}{2}(2x-1)^{\frac{1}{2}} + C$

3. Factorise and simplify the integrated expression into a single term.

$= \frac{1}{6}(2x-1)^{\frac{1}{2}} \left[ (2x-1) + 3 \right] + C$
$= \frac{1}{6}\sqrt{2x-1}(2x+2) + C$
$= \frac{1}{3}(x+1)\sqrt{2x-1} + C$

1. Relate the integral's terms to the substitution $t = \tan x$ by applying algebraic manipulation.

Let $t = \tan x$. Differentiating with respect to $x$ gives:
$\frac{dt}{dx} = \sec^2 x \implies dt = \sec^2 x \,dx$
To introduce these terms into the integral, divide the numerator and the denominator by $\cos^2 x$:
$\int \frac{1}{4\cos^{2}x + 9\sin^{2}x}\,dx = \int \frac{\frac{1}{\cos^2 x}}{4 + 9\frac{\sin^2 x}{\cos^2 x}}\,dx = \int \frac{\sec^2 x}{4 + 9\tan^2 x}\,dx$

2. Express the integral entirely in terms of the new variable $t$.

Substitute $t = \tan x$ and $dt = \sec^2 x \,dx$ into the integral:
$\int \frac{\sec^2 x}{4 + 9\tan^2 x}\,dx = \int \frac{1}{4 + 9t^2}\,dt$

3. Evaluate the integral in terms of $t$ using the standard integration formula for inverse tangent.

Using the standard result $\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$:
$\int \frac{1}{2^2 + (3t)^2}\,dt = \frac{1}{3} \left[ \frac{1}{2} \tan^{-1}\left(\frac{3t}{2}\right) \right] + C$
$= \frac{1}{6} \tan^{-1}\left(\frac{3t}{2}\right) + C$

4. Express the final answer in terms of the original variable $x$.

Replace $t$ with $\tan x$ to obtain the final answer:
$\frac{1}{6} \tan^{-1}\left(\frac{3\tan x}{2}\right) + C$

1. Set up the total rectangle area as a sum

Each rectangle has width $\tfrac1n$ and height equal to the curve value at the right endpoint $x=\tfrac{r}{n}$, for $r=1,2,\dots,n-1$:
$A=\sum_{r=1}^{n-1}\frac1n\left[\left(\frac{r}{n}\right)^{2}+3\left(\frac{r}{n}\right)\right]=\frac1n\left[\frac{1}{n^{2}}\sum_{r=1}^{n-1}r^{2}+\frac{3}{n}\sum_{r=1}^{n-1}r\right].$

2. Substitute the sum formulae

Using the given result (with upper limit $n-1$): $\sum_{r=1}^{n-1}r^{2}=\tfrac16(n-1)n(2n-1)$, and $\sum_{r=1}^{n-1}r=\tfrac12(n-1)n$:
$A=\frac1n\left[\frac{1}{n^{2}}\cdot\frac{(n-1)n(2n-1)}{6}+\frac{3}{n}\cdot\frac{(n-1)n}{2}\right]=\frac1n\left[\frac{(n-1)(2n-1)}{6n}+\frac{3(n-1)}{2}\right].$

3. Simplify to the required closed form

Factor out $(n-1)$ and combine over a common denominator $6n$:
$A=\frac{(n-1)}{n}\left[\frac{2n-1}{6n}+\frac{3}{2}\right]=\frac{(n-1)}{n}\cdot\frac{(2n-1)+9n}{6n}=\frac{(n-1)(11n-1)}{6n^{2}}.$
Hence $A=\dfrac{(n-1)(11n-1)}{6n^{2}}$ as required.

4. Deduce the exact integral by letting n → ∞

As $n\to\infty$ the rectangle width $\tfrac1n\to0$, so the total rectangle area $A$ tends to the exact area under the curve. Therefore
$\int_{0}^{1}\!\left(x^{2}+3x\right)dx=\lim_{n\to\infty}A=\lim_{n\to\infty}\frac{(n-1)(11n-1)}{6n^{2}}=\lim_{n\to\infty}\frac{11n^{2}-12n+1}{6n^{2}}=\lim_{n\to\infty}\left(\frac{11}{6}-\frac{2}{n}+\frac{1}{6n^{2}}\right)=\frac{11}{6}.$