Question:
Integral X + 1 by square root of 2 x - 1 dx
Answer:
Let I = ∫(x + 1)/√(2x - 1) dx ............1
Put 2x - 1 = t2
=> 2dx = 2t*dt
=> dx = t*dt
From equation 1, we get
I = ∫[{(t2 + 1)/2 + 1}/√t2 ]t*dt
=> I = ∫[{(t2 + 1)/2 + 1}/t ]t*dt
=> I = ∫{(t2 + 1)/2 + 1}dt
=> I = ∫{(t2 + 1 + 2)/2}dt
=> I = (1/2)*∫{(t2 + 3) + C
=> I = (1/2)*{(2x - 1)3/2 /3) + √(2x - 1)} + C
=> I = (2x - 1)3/2 /6) + {√(2x - 1)}/2 + C
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