计算int dfrac (dx)(cos x(1+sin x))

步骤 1：代换
将原积分中的$\cos x$替换为$\cos x$，并利用$\cos^2 x = 1 - \sin^2 x$，得到
$$\int \dfrac {dx}{\cos x(1+\sin x)} = \int \dfrac {\cos x}{{\cos }^{2}x(1+\sin x)}dx = \int \dfrac {\cos x}{(1-{\sin }^{2}x)(1+\sin x)}dx$$
步骤 2：换元
令$u = \sin x$，则$du = \cos x dx$，代入上式，得到
$$\int \dfrac {\cos x}{(1-{\sin }^{2}x)(1+\sin x)}dx = \int \dfrac {1}{(1-{u}^{2})(1+u)}du$$
步骤 3：部分分式分解
将$\dfrac {1}{(1-{u}^{2})(1+u)}$分解为部分分式，得到
$$\dfrac {1}{(1-{u}^{2})(1+u)} = \dfrac {1}{2}\left(\dfrac {u}{1-{u}^{2}} + \dfrac {1}{1+u} + \dfrac {1}{{(1+u)}^{2}}\right)$$
步骤 4：积分
分别对每个部分分式进行积分，得到
$$\int \dfrac {1}{(1-{u}^{2})(1+u)}du = \dfrac {1}{2}\int \dfrac {u}{1-{u}^{2}}du + \dfrac {1}{2}\int \dfrac {1}{1+u}du + \dfrac {1}{2}\int \dfrac {1}{{(1+u)}^{2}}du$$
步骤 5：计算积分
计算每个积分，得到
$$\int \dfrac {u}{1-{u}^{2}}du = -\dfrac {1}{2}\ln |1-{u}^{2}|$$
$$\int \dfrac {1}{1+u}du = \ln |1+u|$$
$$\int \dfrac {1}{{(1+u)}^{2}}du = -\dfrac {1}{1+u}$$
步骤 6：合并结果
将每个积分的结果合并，得到
$$\int \dfrac {1}{(1-{u}^{2})(1+u)}du = -\dfrac {1}{4}\ln |1-{u}^{2}| + \dfrac {1}{2}\ln |1+u| - \dfrac {1}{2(1+u)} + C$$
步骤 7：回代
将$u = \sin x$代回，得到
$$\int \dfrac {dx}{\cos x(1+\sin x)} = -\dfrac {1}{4}\ln |1-\sin^2 x| + \dfrac {1}{2}\ln |1+\sin x| - \dfrac {1}{2(1+\sin x)} + C$$
步骤 8：简化
利用$\cos^2 x = 1 - \sin^2 x$，得到
$$\int \dfrac {dx}{\cos x(1+\sin x)} = \dfrac {1}{2}\ln |\dfrac {1+\sin x}{\sqrt {1-\sin^2 x}}| - \dfrac {1}{2(1+\sin x)} + C$$