Question

If \(\int \frac{3x+7}{x^2-3x+2}\,dx = mlog(\frac{x-2}{x-1})+nlog(x-2)+c\), where \(m,n\in R\) and \(c\) is an integration constant, then \(m+n =\)

• A. \(6\)
• B. \(7\)
• C. \(3\)
• D. \(13\)

Hint:

Use sin^6 + cos^6 = 1 - 3 sin^2 cos^2.

Answer 1

The Correct Option is D

Step 1: Simplify the numerator:
\(\sin^6x + \cos^6x = (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos^2x) = 1 - 3\sin^2x\cos^2x\).

Step 2: Split the fraction:
\[ \frac{\sin^6x + \cos^6x}{\sin^2x\cos^2x} = \frac{1}{\sin^2x\cos^2x} - 3 = \sec^2x + \operatorname{cosec}^2x - 3 \]
because \(\frac{1}{\sin^2x\cos^2x} = \frac{\sin^2x + \cos^2x}{\sin^2x\cos^2x} = \sec^2x + \operatorname{cosec}^2x\).

Step 3: Integrate:
\[ \int(\sec^2x + \operatorname{cosec}^2x - 3)\,dx = \tan x - \cot x - 3x + c \]

Final Answer:
The integral is \(\tan x - \cot x - 3x + c\), option (B). \[ \boxed{\tan x - \cot x - 3x + c} \]