Question

int 1-x
upon 2x+3 dx

Answer 1

Step-by-step explanation:

Proof:

Let I = ∫ dx/(2x - 3) …..………………………………………….(1)

Here x is the independent variable. To evaluate the integral in (1), we take recourse to the substitution method.

Put 2x - 3 = y ………………………………….…………………(2)

Taking differentials,

2. dx - 0 = dy

⇒ dx = dy/2

Substituting for dx and (2x-3) in (1),

I = ∫dy/2y = (1/2) ∫dy/y = (1/2) . log y + C

Now revert to the original variable x using (2).

∴ I = (1/2) . log (2x - 3) + C where C = constant of integration.

Hence integral of 1/(2x - 3) dx = [log (2x - 3)]/2 + C (Answer)

Answer Check:

dI/dx = d/dx[(1/2).log (2x-3) + C] = (1/2) .d/dx[log (2x-3)+C]

= (1/2) .d/dx[log (2x-3)]+0] = (1/2).2.1/ (2x-3) = 1/(2x - 3) = Integrand

Hence answer is correct.

Answer 2

Step-by-step explanation:

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