本题中你必须写出解题过程的所有步骤。

完全依赖计算器技术的解法不被接受。

(i) 令 $u=e^x-3$，证明

其中 $a,b$ 为待求常数。

(ii) 用积分证明

其中 $p,q$ 为待求常数，$c$ 为任意常数。

按题目指定令 $u=e^x-3$。需要同时改写 $e^x$、$\mathrm{d}x$ 和上下限；换元后被积函数可展开为 $4u+24+\frac{36}{u}$，于是积分结果自然包含 $\ln u$。

Let

Then

and

so

The limits become

and

Therefore,

Integrating term by term gives the primitive

Since $2\leq u\leq4$, the logarithm is defined throughout the integration interval. The integral is therefore $H(4)-H(2)$, where

Hence

Thus

Alternatively, after integrating in $u$, substitute $u=e^x-3$ back into $H$:

Then $G(\ln7)=128+36\ln4$ and $G(\ln5)=56+36\ln2$. Using the original limits gives the same result:

连续使用两次分部积分。第一次会产生 $\int e^x\sin2x\,\mathrm{d}x$；第二次处理这个积分后，原积分会再次出现。把它移到等式同一边，即可解出原积分。

Let

Take $u=3e^x$ and $\mathrm{d}v=\cos2x\,\mathrm{d}x$, so $\mathrm{d}u=3e^x\,\mathrm{d}x$ and $v=\frac12\sin2x$. Integration by parts gives

Write the remaining integral as

For this second integration by parts, take $u=e^x$ and $\mathrm{d}v=\sin2x\,\mathrm{d}x$, so $v=-\frac12\cos2x$. Then

Substituting this into $I=\frac32e^x\sin2x-\frac32J$,

Therefore,

Dividing by $\frac54$ and including the constant of integration,

Hence

把 $\cos2x$ 选为求导因子，把 $3e^x$ 选为积分因子。第一次分部积分得到含 $\sin2x$ 的积分；第二次仍对指数因子积分，原积分就会以 $-4I$ 的形式出现。移项得到 $5I$，即可求出两个系数。

Let

Take $u=\cos2x$ and $\mathrm{d}v=3e^x\,\mathrm{d}x$, so $\mathrm{d}u=-2\sin2x\,\mathrm{d}x$ and $v=3e^x$. Integration by parts gives

Let

This time take $u=\sin2x$ and $\mathrm{d}v=e^x\,\mathrm{d}x$, so $\mathrm{d}u=2\cos2x\,\mathrm{d}x$ and $v=e^x$. A second integration by parts gives

Substitute this into $I=3e^x\cos2x+6J$:

Collecting the terms in $I$,

Dividing by $5$ and including the constant of integration,

Hence