题目 09
已知：
$$
f^{\prime}(x)=\arctan \left[(x-1)^{2}\right], \quad f(0)=0 \Rightarrow
$$
则：
$$
I=\int_{0}^{1} f(x) \mathrm{~ d} x = ?
$$
解析 09
把“未知”往“已知”形式上凑：
$$
I=\int_{0}^{1} f(x) \mathrm{~ d} x=\int_{0}^{1} f(x) \mathrm{~ d} (x-1)=
$$
$$
\left.(x-1) f(x)\right|_{0} ^{1}-\int_{0}^{1}(x-1) f^{\prime}(x) \mathrm{~ d} x \Rightarrow
$$
$$
I=-\int_{0}^{1}(x-1) \arctan \left[(x-1)^{2}\right] \mathrm{~ d} (x-1)
$$
又：
$$
{\left[(x-1)^{2}\right]^{\prime}=2(x-1) \Rightarrow}
$$
于是：
$$
I=\frac{-1}{2} \int_{0}^{1} \arctan \left[(x-1)^{2}\right] \mathrm{~ d} \left[(x-1)^{2}\right] \Rightarrow
$$
$$
u=(x-1)^{2} \Rightarrow x \in(0,1) \Rightarrow u \in(1,0) \Rightarrow
$$
$$
I=\frac{-1}{2} \int_{1}^{0} \arctan u \mathrm{~ d} u \Rightarrow
$$
$$
I=\frac{1}{2} \int_{0}^{1} \arctan u \mathrm{~ d} u \Rightarrow
$$
$$
\frac{1}{2}\left[\left.u \arctan u\right|_{0} ^{1}-\int_{0}^{1} u \cdot \frac{1}{1+u^{2}} \mathrm{~ d} u\right] \Rightarrow
$$
$$
\frac{1}{2}\left[\frac{\pi}{4}-\left.\frac{1}{2} \ln \left(u^{2}+1\right)\right|_{0} ^{1}\right] \Rightarrow
$$
$$
I=\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2} \ln 2\right] \Rightarrow I=\frac{\pi}{8}-\frac{1}{4} \ln 2
$$