Question

If \(a,b,c,d\neq 0\) and \[ f\left(\frac{at+b}{ct+d}\right)=t, \] then \[ \int f(x)\,dx= \] is:

• A. \(-\dfrac{d}{c}x+\dfrac{bc-ad}{c^2}\log(cx-a)+K\)
• B. \(-\dfrac{d}{c}x+\dfrac{bc-ad}{c^2}\log(cx+d)+K\)
• C. \(\dfrac{a}{c}x+\dfrac{ad-bc}{c^2}\log(cx+a)+K\)
• D. \(\dfrac{a}{c}x-\dfrac{ad-bc}{c}\log(cx+a)+K\)

Hint:

For functional equations of the form \[ f\left(\frac{at+b}{ct+d}\right)=t, \] first put \[ x=\frac{at+b}{ct+d} \] and solve for \(t\) in terms of \(x\).

Answer 1

The Correct Option is A

Step 1: Put the expression equal to \(x\).
Given, \[ f\left(\frac{at+b}{ct+d}\right)=t \] Let \[ x=\frac{at+b}{ct+d} \] Now solve for \(t\).
\[ x(ct+d)=at+b \] \[ cxt+dx=at+b \] \[ cxt-at=b-dx \] \[ t(cx-a)=b-dx \] Therefore, \[ t=\frac{b-dx}{cx-a} \] Hence, \[ f(x)=\frac{b-dx}{cx-a} \]

Step 2: Rewrite \(f(x)\) for integration.
Now, \[ f(x)=\frac{b-dx}{cx-a} \] Divide the expression: \[ \frac{b-dx}{cx-a} = -\frac{d}{c}+\frac{bc-ad}{c(cx-a)} \] Thus, \[ f(x)=-\frac{d}{c}+\frac{bc-ad}{c(cx-a)} \]

Step 3: Integrate \(f(x)\).
\[ \int f(x)\,dx = \int \left(-\frac{d}{c}+\frac{bc-ad}{c(cx-a)}\right)\,dx \] \[ = -\frac{d}{c}x+\frac{bc-ad}{c}\int \frac{dx}{cx-a} \] Since \[ \int \frac{dx}{cx-a}=\frac{1}{c}\log(cx-a), \] we get \[ \int f(x)\,dx = -\frac{d}{c}x+\frac{bc-ad}{c}\cdot \frac{1}{c}\log(cx-a)+K \] \[ = -\frac{d}{c}x+\frac{bc-ad}{c^2}\log(cx-a)+K \]

Step 4: Final conclusion.
Hence, \[ \boxed{ -\frac{d}{c}x+\frac{bc-ad}{c^2}\log(cx-a)+K } \]