calculus - $int frac{sin x}{x}dx=sin x log x$?

I know that integration of $\int \frac{\sin x}{x}dx=\text{Si} (x)$, but when I tried it myself I am getting a different answer. Please check where I am going wrong.

To find:
Closed form of $\int \dfrac{\sin x}{x}dx$

Integrating by parts:

$$\int \frac{\sin x}{x}dx=\sin x \log x- \int \cos x \log x dx$$

Now finding: $ \int \cos x \log x$

$$ \int \cos x \log x=-\log x \sin x - \int \frac{-\sin x}x dx$$

Putting value of $ \int \cos x \log x$ back:

$$\int \frac{\sin x}{x}dx = \sin x \log x- \left(-\log x \sin x - \int \frac{-\sin x}x dx\right)$$

$$2\int \frac{\sin x}{x}dx=2\sin x \log x$$

$$\int \frac{\sin x}{x}dx=\sin x \log x$$

Answer

In your second integration by parts, you got the signs wrong (and you forgot a $dx$). It should be
$$\int \cos x \log x dx = \log x \sin x - \int \frac{\sin x}{x} dx$$
If you plug this back into your first formula, you get:
$$\int \frac{\sin x}{x} dx = \log x \sin x - \left( \log x \sin x - \int \frac{\sin x}{x} dx \right) = \int \frac{\sin x}{x} dx$$
which doesn't say much. Basically you did the same integration by parts twice, once in one direction then in reverse. You cannot compute anything like that.