For indefinite integral, ∫ DX / (11 + 5x) ^ 3,

First, DX = 1 / 5 d (5x + 11)
∫dx/(11+5x)^3
= 1/5 ∫d(5x+11) / (11+5x)^3
= -1 / [10(11+5x)^2] + C

Excuse me, when we use the second substitution method to find the indefinite integral ∫ (a ^ 2-x ^ 2) DX, a > 0, we can make x = asint, (- π / 2 ≤ t ≤ π / 2), and get √ (a ^ 2-x ^ 2) = a √ (1-sin ^ 2t) = acoxt, DX = acostdt
Excuse me, since we have made x = asint, especially - π / 2 ≤ t ≤ π / 2, it does not mean that there is cost

Let x = asint, - π / 2 ≤ t ≤ π / 2, then cost & gt; 0, in addition, the square root must be positive, just like the root 9 = 3, not plus or minus 3

= 1/2 ∫ x^2*e^(x^2)*dx^2
= 1/2 [ ∫ t*e^t dt ]
=1/2 [ t*e^t -e^t ]
=1 / 2 * (x ^ 2-1) * e ^ (x ^ 2)