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I'm not sure how to approach this question, but I know it has something to do with taking the derivative of this. Can anyone please help me out?

Let x > 0. Prove that the value of the following expression does not depend on x: $$\int_{0}^{x}\frac{1}{1+t^4} dt + \frac{1}{3}\int_{0}^\frac{1}{x^3} \frac{1}{1+t^\frac{4}{3}} dt$$

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    $\begingroup$ You told yourself how to do this: take the derivative (using the fundamental theorem of calculus, and, for the second integral, the chain rule). Something you find out about the derivative should tell you the sum is constant. $\endgroup$ Commented Feb 10, 2018 at 20:33

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Let $$f(x) := \int_0^x \frac{1}{1+t^4} dt + \frac{1}{3}\int_0^{\frac{1}{x^3}} \frac{1}{1+t^{\frac{4}{3}}} dt$$ Then differentiating, $$f'(x) = \frac{1}{1+x^4} + \frac{1}{3} \left( \frac{1}{1+x^{-4}} \cdot \frac{-3}{x^4} \right) = \frac{1}{1+x^4} - \left( \frac{1}{x^{4} + 1} \right) = 0$$ Hence, since $f'(x) = 0$ for all $x$, $f$ must be constant, so it does not depend on $x$.

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  • $\begingroup$ $\frac{1}{x^3} = x^{-3}$. So, $\frac{1}{1+(x^{-3})^{\frac{4}{3}}} = \frac{1}{1+x^{-4}}$. $\endgroup$
    – James Yang
    Commented Feb 10, 2018 at 20:50
  • $\begingroup$ What do you mean by f must be constant? $\endgroup$
    – winonator
    Commented Feb 10, 2018 at 20:50
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    $\begingroup$ If $f'(x) = 0$ for all $x$, then $f(x) = c$ for some constant c. One can prove this using Mean Value Theorem. $\endgroup$
    – James Yang
    Commented Feb 10, 2018 at 20:51
  • $\begingroup$ Also isn't the second integral supposed to be $\frac{1}{x^4}$ after you multiply it by $\frac{-3}{x^4}$? $\endgroup$
    – winonator
    Commented Feb 10, 2018 at 20:55
  • $\begingroup$ I added one more step to clarify. $\endgroup$
    – James Yang
    Commented Feb 10, 2018 at 20:56

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