Answer
$\int\frac{cosx-1}{x}dx=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$
Work Step by Step
$\frac{cosx-1}{x}=-\frac{1}{x}+\frac{cosx}{x}$
$=-\frac{1}{2!}x+\frac{1}{4!}x^{3}-\frac{1}{6!}x^{5}+......+\frac{(-1)^{n}x^{2n-1}}{2n!}x+...$
Now,
$\int\frac{cosx-1}{x} dx=\int(-\frac{1}{2!}x+\frac{1}{4!}x^{3}-\frac{1}{6!}x^{5}+...+\frac{(-1)^{n}x^{2n-1}}{2n!}x+..)dx$
$=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$
Hence, $\int\frac{cosx-1}{x}dx=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$
