If int left( frac{4 e^x + 6 e^{-x}}{9 e^x - 4 | vedclass

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

Question

(C) Let $I = \int \frac{4 e^x + 6 e^{-x}}{9 e^x - 4 e^{-x}} d x = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} d x$. We express the numerator as $4 e^{2 x

Vedclass · Accepted Answer

$(-\frac{3}{2}, \frac{35}{36})$

Vedclass · Answer

(C) Let $I = \int \frac{4 e^x + 6 e^{-x}}{9 e^x - 4 e^{-x}} d x = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} d x$. We express the numerator as $4 e^{2 x} + 6 = A(9 e^{2 x} - 4) + B \frac{d}{d x}(9 e^{2 x} - 4)$. $4 e^{2 x} + 6 = A(9 e^{2 x} - 4) + B(18 e^{2 x})$. Equating the coefficients of $e^{2 x}$ and the constant terms: $9 A + 18 B = 4$ and $-4 A = 6$. From $-4 A = 6$,we get $A = -\frac{3}{2}$. Substituting $A$ into the first equation: $9(-\frac{3}{2}) + 18 B = 4 \Rightarrow -\frac{27}{2} + 18 B = 4 \Rightarrow 18 B = 4 + \frac{27}{2} = \frac{35}{2} \Rightarrow B = \frac{35}{36}$. Thus,$I = \int \left( A + B \frac{18 e^{2 x}}{9 e^{2 x} - 4} \right) d x = A x + B \log |9 e^{2 x} - 4| + C$. Therefore,$(A, B) = (-\frac{3}{2}, \frac{35}{36})$.

If $\int \left( \frac{4 e^x + 6 e^{-x}}{9 e^x - 4 e^{-x}} \right) d x = A x + B \log |9 e^{2 x} - 4| + C$,then $(A, B) = $

Similar Questions

Integrate the function: $\frac{5 x-2}{1+2 x+3 x^{2}}$

Let $f(x) = \int \frac{dx}{(3+4x^2) \sqrt{4-3x^2}}$,$|x| < \frac{2}{\sqrt{3}}$. If $f(0) = 0$ and $f(1) = \frac{1}{\alpha \beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$,where $\alpha, \beta > 0$,then $\alpha^2 + \beta^2$ is equal to $.........$.

$\int \sqrt{\frac{a+x}{a-x}} \, dx = $

The value of the integral $\int_0^1 \frac{x^b - 1}{\log x} \, dx$ is

If $\int \frac{1}{1-\cot x} dx = Ax + B \log |\sin x - \cos x| + C$,then $A + B = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module