JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(f\left( {\frac{{x - 4}}{{x + 2}}} \right) = 2x + 1,(x \in R = \left\{ {1, - 2} \right\}),\) then \(\int {f(x)} \,dx\) is equal to (where \(C\) is a constant of integration)
- A \(12\,{\log _e}\left| {1 - x} \right| - 3x + c\)
- B \(-12\,{\log _e}\left| {1 - x} \right| - 3x + c\)
- C \(-12\,{\log _e}\left| {1 - x} \right| + 3x + c\)
- D \(12\,{\log _e}\left| {1 - x} \right| + 3x + c\)
Answer & Solution
Correct Answer
(B) \(-12\,{\log _e}\left| {1 - x} \right| - 3x + c\)
Step-by-step Solution
Detailed explanation
\(\text { (b) Suppose, } \frac{x-4}{x+2}=y\) \( \Rightarrow x-4=y(x+2)\) \(\Rightarrow x(1-y)=2 y+4\) \( \Rightarrow x=\frac{2 y+4}{1-y}\) \(\text { So, } f(y)=2\left(\frac{2 y+4}{1-y}\right)+1\) \(\text { Now, } f(x)=2\left(\frac{2 x+4}{1-x}\right)+1\) \(=\frac{3 x+9}{1-x}\)…
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