Question

If \[f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = x + 2\], then \[\int {f(x)} dx = \]
A. \[{e^{x + 2}}\log |\dfrac{{3x - 4}}{{3x + 4}}| + c\]
B. \[ - \dfrac{8}{3}\log |1 - x| + \dfrac{2}{3}x + c\]
C. \[\dfrac{8}{3}\log |1 - x| + \dfrac{x}{3} + c\]
D. \[{e^{\dfrac{{3x - 4}}{{3x + 4}}}} - \dfrac{{{x^2}}}{2} - 2x + c\]

Answer 1

Hint: First of all, we need to find the value of \[f(x)\]. For that, we will first let \[\dfrac{{3x - 4}}{{3x + 4}}\] be equal to some variable \[t\]. After that, we will apply componendo and dividendo rule on \[\dfrac{{3x - 4}}{{3x + 4}} = t\] and then find the value of \[x\] in terms of \[t\]. After that, we will find the value of \[x + 2\] in terms of \[t\]. After finding the value of \[x + 2\] in terms of \[t\], we will substitute the value of \[\dfrac{{3x - 4}}{{3x + 4}}\] and \[x + 2\] in the given condition and then obtain a function \[f(t)\] which will be in terms of \[t\] only. After we obtain the value of \[f(t)\], we can replace \[t\] by \[x\] and then we will have our \[f(x)\]. Then, we will just integrate the value of \[f(x)\] with respect to \[x\] and we will obtain the required answer.

Formula used:
COMPONENDO AND DIVIDENDO RULE:
\[\dfrac{a}{b} = \dfrac{c}{d} \Leftrightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\]

Complete step by step answer:
We have, \[f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = x + 2 - - - - - (1)\]
Let us suppose,
\[\dfrac{{3x - 4}}{{3x + 4}} = t - - - - - (2)\]
\[ \Rightarrow \dfrac{{3x - 4}}{{3x + 4}} = \dfrac{t}{1}\]
Applying componendo and dividendo, we get
\[ \Rightarrow \dfrac{{(3x - 4) + (3x + 4)}}{{(3x - 4) - (3x + 4)}} = \dfrac{{t + 1}}{{t - 1}}\]
\[ \Rightarrow \dfrac{{3x - 4 + 3x + 4}}{{3x - 4 - 3x - 4}} = \dfrac{{t + 1}}{{t - 1}}\]
Adding the terms in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{6x}}{{ - 8}} = \dfrac{{t + 1}}{{t - 1}}\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow - \dfrac{{3x}}{4} = \dfrac{{t + 1}}{{t - 1}}\]
\[ \Rightarrow - \dfrac{3}{4}x = \dfrac{{t + 1}}{{t - 1}}\]

By cross Multiplying
\[ \Rightarrow x = - \dfrac{4}{3}\left( {\dfrac{{t + 1}}{{t - 1}}} \right)\]
Now, adding 2 both sides.
\[ \Rightarrow x + 2 = - \dfrac{4}{3}\left( {\dfrac{{t + 1}}{{t - 1}}} \right) + 2\]
\[ \Rightarrow x + 2 = \dfrac{{ - 4(t + 1)}}{{3(t - 1)}} + 2\]
Taking LCM on the right hand side
\[ \Rightarrow x + 2 = \dfrac{{ - 4(t + 1) + (2 \times 3(t - 1))}}{{3(t - 1)}}\]
\[ \Rightarrow x + 2 = \dfrac{{ - 4(t + 1) + 6(t - 1)}}{{3(t - 1)}}\]
Now opening the brackets, we get
\[ \Rightarrow x + 2 = \dfrac{{ - 4t - 4 + 6t - 6}}{{3(t - 1)}}\]

Taking variables together in numerator,
\[ \Rightarrow x + 2 = \dfrac{{( - 4t + 6t) - 4 - 6}}{{3(t - 1)}}\]
\[ \Rightarrow x + 2 = \dfrac{{2t - 10}}{{3(t - 1)}}\]
Taking out \[2\]common from the numerator on the right side.
\[ \Rightarrow x + 2 = \dfrac{{2(t - 5)}}{{3(t - 1)}} - - - - - (3)\]
Now substituting (2) and (3) in (1), we get,
\[ \Rightarrow f(t) = \dfrac{{2(t - 5)}}{{3(t - 1)}}\]
Now, replacing \[t\] by \[x\] in the above equation, we get
\[ \Rightarrow f(x) = \dfrac{{2(x - 5)}}{{3(x - 1)}}\]
We need to find \[\int {f(x)} dx\] which means, we need to find \[\int {\dfrac{{2(x - 5)}}{{3(x - 1)}}} dx\]

So now, we will solve \[\int {\dfrac{{2(x - 5)}}{{3(x - 1)}}} dx\]
\[ \Rightarrow \int {\dfrac{{2(x - 5)}}{{3(x - 1)}}} dx = \dfrac{2}{3}\int {\dfrac{{x - 5}}{{x - 1}}dx} \]
As we can write \[ - 5 = - 4 - 1\], using this we get
\[ \Rightarrow \dfrac{2}{3}\int {\dfrac{{x - 4 - 1}}{{x - 1}}dx} \]
\[ \Rightarrow \dfrac{2}{3}\int {\dfrac{{(x - 1) - 4}}{{x - 1}}} dx\]
Splitting the denominator, we have
\[\Rightarrow \dfrac{2}{3}\int {\left[ {\dfrac{{(x - 1)}}{{x - 1}} - \dfrac{4}{{x - 1}}} \right]} dx\]
Cancelling common factors in numerator and denominator and separating the integrals, we get,
\[\Rightarrow \dfrac{2}{3}\int {\left[ {1 - \dfrac{4}{{x - 1}}} \right]} dx\]
\[\Rightarrow \dfrac{2}{3}\left[ {\int 1 dx - \int {\dfrac{4}{{x - 1}}dx} } \right]\]

Using the integration formula\[\int 1 dx = x + c\]
\[\Rightarrow \dfrac{2}{3}\left[ {x - 4\int {\dfrac{{dx}}{{x - 1}}} } \right]\]
Using \[\int {\dfrac{{dx}}{{x + c}} = \log |x + c| + k} \], we have
\[\Rightarrow \dfrac{2}{3}[x - 4\log |x - 1|] + c\]
Opening the brackets, we get
\[\Rightarrow \dfrac{2}{3}x - \dfrac{2}{3} \times 4\log |x - 1| + c\]
\[\Rightarrow \dfrac{2}{3}x - \dfrac{8}{3}\log |x - 1| + c\]
Hence, we get
\[\int {f(x)} dx = \dfrac{2}{3}x - \dfrac{8}{3}\log |x - 1| + c\]
\[\therefore \int {f(x)} dx = - \dfrac{8}{3}\log |1 - x| + \dfrac{2}{3}x + c\]

Therefore, the correct option is B.

Note: To find the value of \[f(x)\], we should keep in mind that we have to apply componendo and dividendo otherwise we will not be able to solve the problem for \[x\] in terms of \[t\]. Now, after Finding the value of \[f(x)\], we need to integrate \[f(x)\]. We might get stuck while solving the integration value. So, we need to keep in mind the method to solve the integration. For such types of integration, we have to split some values from the numerator and then split the denominator and cancel some of the terms from numerator and denominator and then arrive at such an expression which can be integrated easily.