Question in relation to Fundamental Theorem of Calculus

Question

I obtain two answers, one is $\dfrac1{\sqrt{1+x^6}}$ and another one is $\dfrac{2x}{\sqrt{1+x^{12}}}$ by using Fundamental Theorem of Calculus, but I am not so sure. Would anyone help me?

Answer 1

Let $$ F(x) = \int_0^x \frac{1}{\sqrt{1+t^6}}dt $$ and write $f(x) = F(x^2) - F(-x)$.

Does this help?

Answer 2

Don't forget about the derivatives at the boundaries: $$f(x)=\int_{a(x)}^{b(x)}g(t)dt=F(b(x))-F(a(x))$$ $F$ is the indefinite integral. Derivative: $$df/dx=F'(b(x))b'(x)-F'(a(x))a'(x)=g(b(x))b'(x)-g(a(x))a'(x)$$

$$=\frac{1}{\sqrt{1+(x^2)^6}}(2x)-\frac{1}{\sqrt{1+(-x)^6}}(-1)$$

You see now, you have to combine both your expressions. Both boundaries are important, not just one of them.