From Thomas Calculus we are given the following integral to differentiate
Applying Leibniz Integral rule I got the following result
$-2x\sqrt{x^6+x^2} - \int_1^{x^2}\frac{x}{\sqrt{t^3+x^2}}dt$
Checking my work against the instructor's solution manual I find the solution manual reads
I did not apply exercise 51 to this problem nor did I find it necessary.
I believe the instructor's solution is incorrect. In the left half of the final answer the instructor has $-x^2\sqrt{x^6+x^2}$ however it should be $-2x\sqrt{x^6+x^2}$
I have attempted to find duplicates of this problem solution on the internet to check for this error, but have not found any, nor is there any simple way (that I have attempted) to verify this error computationally.
Can someone please verify my claims?
Making the problem more general,if $$F(x)=\int_{a(x)}^{b(x)} f(t,x) \, dt$$ the fundamantal theorem of calculus gives $$F'(x)=b'(x) f(b(x),x)-a'(x) f(a(x),x)+\int_{a(x)}^{b(x)}\frac{\partial f(t,x)}{\partial x} \, dt$$ So, for your case $$F'(x)=-2 x \sqrt{x^6+x^2}+x\int_{x^2}^1 \frac{dt}{\sqrt{t^3+x^2}}$$
SO, your work is perfectly correct and then $\to +1$.