∫(1 + x)/(√1 – x2) dx
u = √1 – x2
du = -2x(1/2)[1/(√1 – x2)] dx = [- x/ (√1 – x2)] dx
dx = [- (√1 – x2)/ x] du = [- u/ (√1 – u2)] du
1 – x2 = u2
x = √1 – u2
∫(1 + x)/(√1 – x2) dx = ∫[(1 + √1 – u2)/ u] [- u/ (√1 – u2)] du
∫[-(1 + √1 – u2)/ (√1 – u2)] du = - ∫[(1 + √1 – u2)/ (√1 – u2)] du
- ∫[(1/ √1 – u2)] du - ∫[(√1 – u2) / (√1 – u2)] du =
- arcsen u + c – (u + c)
- arcsen u – u + c
∫(1 + x)/(√1 – x2) dx = - arcsen (√1 – x2) – (√1 – x2) + c
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