If fleft(frac{2 x+3}{3 x+5}right)=x+4, where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$. Let $y = \frac{2 x+3}{3 x+5}$.Then $y(3x+5) = 2x+3 \Rightarrow 3xy + 5y = 2x+3 \Rightarrow x(3y-2)

Vedclass · Accepted Answer

$\frac{-8}{3}$

Vedclass · Answer

(D) Given $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$. Let $y = \frac{2 x+3}{3 x+5}$.Then $y(3x+5) = 2x+3 \Rightarrow 3xy + 5y = 2x+3 \Rightarrow x(3y-2) = 3-5y \Rightarrow x = \frac{3-5y}{3y-2}$.Substituting $x$ in $f(y) = x+4$, we get $f(y) = \frac{3-5y}{3y-2} + 4 = \frac{3-5y + 12y - 8}{3y-2} = \frac{7y-5}{3y-2}$.Thus, $f(x) = \frac{7x-5}{3x-2}$.Now, $\int f(x) dx = \int \frac{7x-5}{3x-2} dx = \int \frac{\frac{7}{3}(3x-2) - 5 + \frac{14}{3}}{3x-2} dx = \int \left( \frac{7}{3} - \frac{1/3}{3x-2} \right) dx$.$= \frac{7}{3}x - \frac{1}{9} \ln |3x-2| + C$.Comparing with $Ax + B \ln |3x-2| + C$, we get $A = \frac{7}{3}$ and $B = -\frac{1}{9}$.Therefore, $3B - A = 3(-\frac{1}{9}) - \frac{7}{3} = -\frac{1}{3} - \frac{7}{3} = -\frac{8}{3}$.

If $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$, where $x \neq \frac{-5}{3}, \frac{-2}{3}$, and $\int f(x) d x=A x+B \ln |3 x-2|+C$, then $3 B-A=$

Similar Questions

$\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx = $

$\int \frac{1 + x + \sqrt{x + x^2}}{\sqrt{x} + \sqrt{1 + x}} dx =$

Find the following integral: $\int(2x^2 - 3\sin x + 5\sqrt{x}) dx$

If $\int \frac{\cos 3x}{\sin x} dx = p \cos 2x + q \log |\sin x| + C$,then $p + q =$ . . . . . . .

Integrate the function: $\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module