यदि fleft( frac{x - 4}{x + 2} right) = 2x + 1 | vedclass

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

Question

(B) माना $y = \frac{x - 4}{x + 2}$.तब $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.$x$ का म

Vedclass · Accepted Answer

$-12 \log_e |1 - x| - 3x + C$

Vedclass · Answer

(B) माना $y = \frac{x - 4}{x + 2}$.तब $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.$x$ का मान $f(y) = 2x + 1$ में रखने पर,हमें प्राप्त होता है $f(y) = 2\left( \frac{2y + 4}{1 - y} \right) + 1$.$f(y) = \frac{4y + 8 + 1 - y}{1 - y} = \frac{3y + 9}{1 - y}$.अतः,$f(x) = \frac{3x + 9}{1 - x} = \frac{3(x - 1 + 4)}{1 - x} = \frac{3(x - 1)}{1 - x} + \frac{12}{1 - x} = -3 + \frac{12}{1 - x}$.अब,$\int f(x) \,dx = \int \left( -3 + \frac{12}{1 - x} \right) \,dx$.$= -3x + 12 \int \frac{1}{1 - x} \,dx$.$= -3x + 12 \left( \frac{\log_e |1 - x|}{-1} \right) + C$.$= -12 \log_e |1 - x| - 3x + C$.

यदि $f\left( \frac{x - 4}{x + 2} \right) = 2x + 1$ जहाँ $x \in R \setminus \{ -2 \}$ है,तो $\int f(x) \,dx$ का मान ज्ञात कीजिए (जहाँ $C$ समाकलन स्थिरांक है)।

Similar Questions

यदि $\int \frac{x}{(a+x)^5} dx = \frac{1}{k(a+x)^4}(f(x)) + c$ है, तो $\frac{f(-a)}{ak} = $

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

$\int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx} = $

यदि $\int {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} dx = \frac{1}{P}\left( {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} \right) + K$; तो $P =$ ?

$\int (x + \frac{1}{x})^3 dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module