题目

Problem

Find

(i)

\int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x

(2)

(ii)

\int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x \qquad x\ne 0

(2)

题目中文翻译

求

(i)

\int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x

(ii)

\int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x \qquad x\ne 0

解答

展开

分母的导数恰好是分子的两倍，因此可直接使用 $\int f'(x)/f(x)\,\mathrm{d}x=\ln|f(x)|+c$ 。

展开

We need to find

\int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x

Notice that the derivative of the denominator is

\frac{\mathrm{d}}{\mathrm{d}x}(3x^2-4x+5)=6x-4

and

6x-4=2(3x-2)

So the numerator is half the derivative of the denominator.

Therefore

\begin{align*} \int \frac{3x-2}{3x^2-4x+5}\,\mathrm{d}x =&\,\frac12\int \frac{6x-4}{3x^2-4x+5}\,\mathrm{d}x\\[4mm] =&\,\frac12\ln(3x^2-4x+5)+c \end{align*}

Hence

\boxed{\frac12\ln(3x^2-4x+5)+c}

展开

令 $u=e^{2x}-1$ ，分子中的 $e^{2x}\,\mathrm{d}x$ 随即变成 $\frac12\,\mathrm{d}u$ ，积分化为幂函数形式。

展开

We need to find

\int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x

Let

u=e^{2x}-1

Then

\frac{\mathrm{d}u}{\mathrm{d}x}=2e^{2x}

e^{2x}\,\mathrm{d}x=\frac12\,\mathrm{d}u

Therefore

\begin{align*} \int \frac{e^{2x}}{(e^{2x}-1)^3}\,\mathrm{d}x =&\,\frac12\int u^{-3}\,\mathrm{d}u\\[4mm] =&\,\frac12\cdot \frac{u^{-2}}{-2}\\[4mm] =&\,-\frac14u^{-2} \end{align*}

Substitute back:

\boxed{-\frac{1}{4(e^{2x}-1)^2}+c}