Let f (x) be continuous, D / DX ∫ superscript x subscript 0tf (x ^ 2-T ^ 2) DT =?

Let f (x) be continuous, D / DX ∫ superscript x subscript 0tf (x ^ 2-T ^ 2) DT =?

It's hard for me to find your problem!
Solution 1: exchange method!
Let u = x Λ 2-T Λ 2, then t = √ (x Λ 2-u)
When t = 0, u = x Λ 2; when t = x, u = 0
And DT = - 1 / 2 √ (x Λ 2-u)
The original formula = ∫ f (U) * √ (x ∫ 2-u) * (- 1) / 2 √ (x ∫ 2-u) Du = - 1 / 2 ∫ f (U) Du (upper limit 0, lower limit x ∫ 2) = 1 / 2 ∫ f (U) Du (Upper limit x ∫ 2, lower limit 0)
＝1/2f（x∧2）*2x
＝x＊f(x∧2).
Solution 2
Let u Λ 2 = x Λ 2-T Λ 2
Then 2udu = - 2tdt, х DT = - U / TDU, when t = 0, u = x, when t = x, u = 0
The original formula = ∫ t * f (u∧2) * - U / TDU
= ∫ f (u∧2) * (- U) Du (upper limit 0, lower limit x)
= ∫ U * f (u∧2) Du (upper limit x lower limit 0)
＝x*f（x∧2）.

D / DX ∫ DT / √ 1 + T ^ 4 the superscript is x ^ 3 and the subscript is x ^ 2

I haven't been in touch with mathematics for a long time
We know that D / dt fxdx is superscript U (T) and subscript v (T) equals u '(T) f (U (T)) - V' (T) f (V (T))
Here we use sqrt for square root,
By substituting the above formula, we get 3x ^ 2 / sqrt (1 + x ^ 12) - 2x / sqrt (1 + x ^ 8)
Ah, the symbols are so hard to type