If \( y=\sqrt{x}-\frac{1}{\sqrt{x}} \quad \) prove that \( 2 x \frac{d y}{d x}+y=2 \sqrt{x} \)

\( \int _ { 0 } ^ { x } y ( t ) d t = x _ { + } ^ { 3 } \int _ { 0 } ^ { x } \sin ( x , t ) y ( t ) d t \)

\( \left.\begin{array}{l}\text { 4) Expand the following function into Fourier series } \\ \qquad f(x)=\left\{\begin{array}{cc}1+\frac{2 x}{\pi} & -\pi \leq x \leq \frac{-\pi}{2} \\ -\sin x & \frac{-\pi}{2} \leq x \leq 0 \\ 1-\frac{2 x}{\pi} & 0 \leq x \leq \frac{\pi}{2} \\ +\sin x & \frac{\pi}{2} \leq x \leq \pi\end{array}\right. \\ \qquad f(x)=f(x+2 \pi)\end{array}\right\} \)

\( \int ( x ) a x + 8 ) ^ { 2 } \cos x + 2 x \)

7. \( \frac{d}{d x}\left(x^{3}-3 x^{2}+2 x\right) \)

solve \( \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} \)