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let I
Put x+4= t
Let = t
∴2x dx = dt
Let log |x| = t
∴
Let 1 + log x = t
∴
sin x ⋅ sin (cos x)
sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt
Let
∴ 2adx = dt
Let ax + b = t
⇒ adx = dt
Let
∴ dx = dt
Let 1 + 2x2 = t
∴ 4xdx = dt
Let
∴
Let
∴
Let
∴ 9x2dx = dt
Let log x = t
∴
Let
∴ −8x dx = dt
Let
∴ 2dx = dt
Let
∴ 2xdx = dt
Let
∴
Dividing numerator and denominator by ex, we obtain
Let
∴
Let
∴
Let 2x − 3 = t
∴ 2dx = dt
Let 7 − 4x = t
∴ −4dx = dt
Let
∴
Let
∴
Let
∴
Let
∴
Let sin 2x = t
∴
Let
∴ cos x dx = dt
cot x log sin x
Let log sin x = t
Let 1 + cos x = t
∴ −sin x dx = dt
Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
Divide Numerator and Deminator by cos2x
Put tan x = t
sec2x dx = dt
I =
=
=
I =
Let 1 + log x = t
∴
Let
∴
Let x4 = t
∴ 4x3 dx = dt
Let
∴
From (1), we obtain
equals
Let
∴
Hence, the correct answer is D.
Integrate the question
Let
∴ (2x + 1)dx = dt
equals
A.
B.
C.
D.
Hence, the correct answer is B.
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