实验
红色
红色的积分号
<font color="#dd0000">$\int$红色的积分号</font>
1.
(1) ．
解：
(2) .
解：
\int \frac{x+5}{x^2-6x+13}dx = \int \frac{x+5}{(x-3)^2+4}dx =
\frac{1}{4}\int\frac{x+5}{1+(\frac{x-3}{2})^2}dx =
\frac{2}{4}\int \frac{d\frac{(x-3)}{2}}{1+(\frac{x-3}{2})^2} =
\frac{1}{2}\arctan \frac{x-3}{2}+C
4.
(1) ;
解：
\int\frac{dx}{e^x-e^{-x}}=
\int\frac{e^x dx}{e^{2x}-1}=
\int\frac{de^x}{(e^{x})^2-1}=
\int\frac{de^x}{e^x-1}-\int\frac{de^x}{e^x+1}=
\int\frac{d(e^x-1)}{e^x-1}-\int\frac{d(e^x+1)}{e^x+1}=
=\ln |e^x-1|-\ln |e^x+1|+C=
\ln|\frac{e^x-1}{e^x+1}|+C
(2)
解：
\int\frac{x}{(1-x)^3}dx=
-\int\frac{1-x-1}{(1-x)^3}dx=
-\int\frac{1}{(1-x)^2}dx+\int\frac{1}{(1-x)^3}dx=
\int (1-x)^{-2}d(1-x)+\int (1-x)^{-3}d(1-x)
=\frac{1}{-2+1}(1-x)^{-2+1}+\frac{1}{-3+1}(1-x)^{-3+1}+C=
\frac{-1}{1-x}+\frac{-\frac{1}{2}}{(1-x)^2}
---
## 第四章
---
### *总复习题四*
---
**1.**
(1) $\int x^3 e^x dx =$ $\underline{ }$．
**解：**
$\int x^3 e^x dx = \int x^3 d(e^x) = e^x x^3-\int e^xd(x^3) = e^x x^3-3\int e^x x^2dx$
$\int e^x x^2dx=\int x^2d(e^x)=e^x x^2-\int e^xd(x^2)=e^x x^2-2\int e^x xdx$
$\int e^x xdx = \int xd(e^x) = e^x x-\int e^x dx=e^x x-e^x+C$
$\Rightarrow \int x^3 e^xdx = e^x x^3-3[e^x x^2-2(e^x x-e^x)]+C =$ <font color="#dd00dd">$\underline{e^x x^3-3e^x x^2+6e^x x-6e^x+C}$</font>
---
(2) $\int \frac{x+5}{x^2-6x+13}=$ $\underline{ }$.
**解**：
$$
\int \frac{x+5}{x^2-6x+13}dx = \int \frac{x+5}{(x-3)^2+4}dx =
\frac{1}{4}\int\frac{x+5}{1+(\frac{x-3}{2})^2}dx =
\frac{2}{4}\int \frac{d\frac{(x-3)}{2}}{1+(\frac{x-3}{2})^2} =
\frac{1}{2}\arctan \frac{x-3}{2}+C
$$
$\Rightarrow \int \frac{x+5}{x^2-6x+13}dx =$ <font color="#dd00dd"> $\underline{\frac{1}{2}\arctan \frac{x-3}{2}+C}$</font>
---
**4**.
(1)$\int \frac{dx}{e^x-e^{-x}}$ ;
**解：**
$$
\int\frac{dx}{e^x-e^{-x}}=
\int\frac{e^x dx}{e^{2x}-1}=
\int\frac{de^x}{(e^{x})^2-1}=
\int\frac{de^x}{e^x-1}-\int\frac{de^x}{e^x+1}=
\int\frac{d(e^x-1)}{e^x-1}-\int\frac{d(e^x+1)}{e^x+1}=
$$
$$
=\ln |e^x-1|-\ln |e^x+1|+C=
\ln|\frac{e^x-1}{e^x+1}|+C
$$
---
(2)$\int\frac{x}{(1-x)^3}dx$
**解：**
$$
\int\frac{x}{(1-x)^3}dx=
-\int\frac{1-x-1}{(1-x)^3}dx=
-\int\frac{1}{(1-x)^2}dx+\int\frac{1}{(1-x)^3}dx=
\int (1-x)^{-2}d(1-x)+\int (1-x)^{-3}d(1-x)
$$
$$
=\frac{1}{-2+1}(1-x)^{-2+1}+\frac{1}{-3+1}(1-x)^{-3+1}+C=
\frac{-1}{1-x}+\frac{-\frac{1}{2}}{(1-x)^2}
$$
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