Given:
\(x\int_1^x y(t)\,dt=(x+1)\int_1^x ty(t)\,dt\)

Let
\(A=\int_1^x y(t)\,dt\)
and
\(B=\int_1^x ty(t)\,dt\)

So,
\(xA=(x+1)B\)

Differentiate both sides with respect to \(x\):
\(A+xy=(x+1)xy+B\)

From \(xA=(x+1)B\), we get
\(A=\frac{x+1}{x}B\)

Put this value in the differentiated equation:
\(\frac{x+1}{x}B+xy=(x+1)xy+B\)

\(\frac{B}{x}=x^2y\)

So,
\(B=x^3y\)

Now differentiate again:
\(\frac{d}{dx}(B)=\frac{d}{dx}(x^3y)\)

\(xy=3x^2y+x^3y'\)

\(x^3y'=xy-3x^2y\)

\(y'=\left(\frac{1}{x^2}-\frac{3}{x}\right)y\)

\(\frac{dy}{y}=\left(\frac{1}{x^2}-\frac{3}{x}\right)dx\)

Integrating both sides:
\(\log y=-\frac{1}{x}-3\log x+\log C\)

So,
\(y=\frac{C}{x^3}e^{-1/x}\)

Therefore,
Answer: (c) \(\frac{C}{x^3}e^{-1/x}\)