If int frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d | vedclass

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

Question

(C) Let $I = \int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x$. We express the numerator as $A(7 e^x+3 e^{-x}) + B \frac{d}{dx}(7 e^x+3 e^{-x})$. $3 e^x

Vedclass · Accepted Answer

$\frac{38}{21}$

Vedclass · Answer

(C) Let $I = \int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x$. We express the numerator as $A(7 e^x+3 e^{-x}) + B \frac{d}{dx}(7 e^x+3 e^{-x})$. $3 e^x-7 e^{-x} = A(7 e^x+3 e^{-x}) + B(7 e^x-3 e^{-x})$. Comparing coefficients of $e^x$ and $e^{-x}$: $7A + 7B = 3$ and $3A - 3B = -7$. Solving these,$A = -2/21$ and $B = 19/21$. Thus,$I = \int \left( \frac{-2}{21} + \frac{19}{21} \frac{7 e^x-3 e^{-x}}{7 e^x+3 e^{-x}} ight) dx$. $I = \frac{-2}{21} x + \frac{19}{21} \ln |7 e^x+3 e^{-x}| + C$. $I = \frac{-2}{21} x + \frac{19}{21} \ln |e^x(7+3 e^{-2x})| + C$. $I = \frac{-2}{21} x + \frac{19}{21} (x + \ln |7+3 e^{-2x}|) + C$. $I = \frac{17}{21} x + \frac{19}{21} \ln |3(e^{-2x}+7/3)| + C$. $I = \frac{17}{21} x + \frac{19}{21} \ln |e^{-2x}+7/3| + 	ext{constant}$. Comparing with $Kx + L \ln(e^{-2x}+7/3) + C$,we get $K = 17/21$ and $L = 19/21$. $K+L = 17/21 + 19/21 = 36/21 = 12/7$.

If $\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x=K x+L \log \left(e^{-2 x}+\frac{7}{3}\right)+C$,then $K+L=$

Similar Questions

$\begin{aligned} & \int \frac{x \, dx}{\sqrt[15]{\left(1+x^2\right)^{12}\left(2+x^2\right)^{18}}}=\alpha\left(\frac{1+x^2}{2+x^2}\right)^{1 / n}+C \Rightarrow \\ & \frac{n}{\alpha}= \end{aligned}$

Find $\int \sqrt{x^{2}+2 x+5} \, dx$.

$\int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $

If $\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|+c$ (where $c$ is a constant of integration),then the ordered pair $(\lambda, |f(\theta)|)$ is equal to

$\int \frac{dx}{4\sin^2 x + 5\cos^2 x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module