If f(3x-43x+4) = x+2, x -43, and f(x),dx = A|1-x| + Bx + C, then the ordered pair (A,B) is…

Question

If $f\left(\dfrac{3x-4}{3x+4}ight) = x+2$, $x 
eq -\dfrac{4}{3}$, and $\int f(x)\,dx = A\log|1-x| + Bx + C$, then the ordered pair $(A,B)$ is equal to: (where C is a constant of integration)

Accepted Answer

Given,

f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2,     x  $$ \ne $$ - $${4 \over 3}$$

Let, $${{3x - 4} \over {3x + 4}}$$ = t

$$ \Rightarrow $$ $\,\,\,$3x$-$4 = 3tx + 4t$ \Rightarrow $$ $\,\,\,$$ 3x - 3tx = 4t + 4

$$ \Rightarrow $$ $\,\,\,$x =${{4t + 4} \over {3 - 3t}}$So, f(t) =${{4t + 4} \over {3 - 3t}}$+ 2 =${{10 - 2t} \over {3 - 3t}}$$\therefore\,\,\,$f (x) =${{10 - 2x} \over {3 - 3x}}$$\therefore\,\,\,$$\int {f(x)\,dx} $=$\int {{{2x - 10} \over {3x - 3}}} \,dx$=$\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} } $=${2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}} $=${2 \over 3}$x +${2 \over 3}$log$\left| {x - 1} \right|$$-$${{10} \over 3}$log$\left| {x - 1} \right|$+ C

=${2 \over 3}$x$-$${8 \over 3}$log$\left| {x - 1} \right|$+ C$\therefore\,\,\,$A =$-$${8 \over 3}$and B =${2 \over 3}$

Solutions · 1