[mathjax]
問題
解答
\(\displaystyle\int \displaystyle\frac{x\sin x+\cos x}{x^2}dx \)
\(=\displaystyle\int \biggl(\displaystyle\frac{\sin x}{x}+\displaystyle\frac{\cos x}{x^2}\biggr) dx\)
\(=-\displaystyle\frac{\cos x}{x}-\displaystyle\int \displaystyle\frac{\cos x}{x^2}+\displaystyle\int\displaystyle\frac{\cos x}{x^2}dx \) \(\cdots\) 第一項を部分積分。
\(=-\displaystyle\frac{\cos x}{x}+C\)
