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I am learning Calculus online and this problem stumped me. The solution doesn't really make sense either. I'm starting to think that the user made a typo. Here is the question to simplify:

$$\frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt$$

if $u = x^2$, then $\frac{du}{dx} = 2x$, which means $dx = \frac{du}{2x}$ So we have,

$$-\frac{d}{\frac{du}{2x}}\int_1^u\frac{t^4 + 1}{t^2 + 1}dt$$

And then when I simplify it,

$$-\frac{d}{du}\int_1^u\frac{t^4 + 1}{t^2 + 1}*(2x)dt$$

But this still doesn't get rid of the t's. I have no idea how to replace t with x. Also, I am unable to get rid of $dt$, if I could get rid of $dt$ then I would be able to simplify the equation as the derivative of an integral of $f(x)$ is simply $f(x)$

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    $\begingroup$ Let $F$ be any primitive of $f(t)=\frac{t^4+1}{t^2+1}$. Then $\int_{x^2}^1 f(t)\, dt = F(1)-F(x^2)$. Now you can differentiate the right-hand side. $\endgroup$
    – Siminore
    Mar 18, 2014 at 9:54

2 Answers 2

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So, to evaluate $$ \frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt $$ it suffices to note that if $F$ is a primitive (unique up to a constant) of $\frac{t^4 + 1}{t^2 + 1}$, then the integral becomes $F(1) - F(x^2)$. Differentiating in turn gives the result $$ \frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt = \frac{d}{dx} \left( F(1) - F(x^2) \right) = 0 - F'(x^2) \cdot \frac{d}{d x} (x^2) $$ which is simply $$ - 2x \frac{x^8 + 1}{x^4 + 1} . $$

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    $\begingroup$ How did you calculate F(1) = 0? I tried doing this but I found F(1) = 1, since you take the derivative of $(t^4+1)/(t^2+1)$ which becomes $(4t^3(t^2+1)-2t(t^4+1))/(t^2+1)^2$. Then you sub 1 in for $t$, which gives $(4(2)-2(2))/4$ = $1$. $\endgroup$
    – Kelsey
    Dec 5, 2015 at 23:48
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    $\begingroup$ @Kelsey Ah, I see how that might look confusing - Actually, we don't know what $F$ is, it's the function whose derivative is $\frac{t^4 + 1}{t^2 + 1}$. But, we do know that $F(1)$ is some number - i.e., it is just a constant with respect to the variable $x$. So, when we differentiate a constant with respect to $x$, it becomes $0$. That is, $$\frac{d}{dx} (F(0)) = \frac{d}{dx} ( \text{ some number }) = 0. $$ $\endgroup$
    – izœc
    Dec 9, 2015 at 0:03
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When you calculate the derivative of $$\int_{u(x)}^{v(x)} f(t) dt,$$ what you must realize is that this is actually a compositum function. You have $3$ functions: $F(u,v) = \int_u^v{f(t) dt}$ and the functions $u(x)$ and $v(x)$, giving you. $$\int_{u(x)}^{v(x)} f(t) dt = F(u(x),v(x)).$$ The chain rule for such a function is, \begin{align} \frac{d}{dx}F(u(x),v(x)) &= \frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx} \\&=-f(u(x)) \cdot u'(x) + f(v(x))\cdot v'(x). \end{align}

In your case, $f(t)=\frac{t^4+1}{t^2+1}$, $u(x) = x^2$ and $v(x) = 1$.

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