$f(x)=\int_0^x\sin(t^2-t+x)dt$. Find $f''(x)+f'(x)$ $f(x)=\int_0^x\sin(t^2-t+x)dt$. Find $f''(x)+f'(x)$.
Using leibnitz integral rule,
$$f'(x)=\int_0^x\cos(t^2-t+x)+\sin(x^2)dt$$
$$f''(x)=-\int_0^x\sin(t^2-t+x)dt+2x\cos(x^2)$$
Answer given is $(2x+1)\cos(x^2)$. I wont get this if I add $f''(x)+f'(x)$

P.S. :Since OP is convinced that it is asked to find $f''+f$.
Given $f(x)=\int_0^x\sin(t^2-t+x)dt$.
By Leibniz differentiation under the integral rule,
$$f'(x)=\int_0^x\cos(t^2-t+x)dt+\sin(x^2)$$
$$f''(x)=-\int_0^x\sin(t^2-t+x)dt+(2x+1)\cos(x^2)$$
$$f''(x)=-f(x)+(2x+1)\cos(x^2)$$
$$f''(x)+f(x)=(2x+1)\cos(x^2)$$
