If int frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} d | vedclass

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

Question

(D) Let $I = \int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx$. We express the numerator as $A(4e^x - 5e^{-x}) + B(4e^x + 5e^{-x})$, where $4e^x - 5e^{-x

Vedclass · Accepted Answer

$p = -1/8, q = 7/8$

Vedclass · Answer

(D) Let $I = \int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx$. We express the numerator as $A(4e^x - 5e^{-x}) + B(4e^x + 5e^{-x})$, where $4e^x - 5e^{-x}$ is the derivative of the denominator $4e^x + 5e^{-x}$. Equating coefficients of $e^x$ and $e^{-x}$: $4A + 4B = 3$ $-5A + 5B = -5$ From the second equation, $-A + B = -1$, so $B = A - 1$. Substituting into the first: $4A + 4(A - 1) = 3 \implies 8A - 4 = 3 \implies 8A = 7 \implies A = 7/8$. Then $B = 7/8 - 1 = -1/8$. So, $I = \int \frac{7/8(4e^x - 5e^{-x}) - 1/8(4e^x + 5e^{-x})}{4e^x + 5e^{-x}} dx$. $I = 7/8 \int \frac{4e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx - 1/8 \int 1 dx$. $I = 7/8 \log |4e^x + 5e^{-x}| - 1/8 x + C$. Comparing with $px + q \log |4e^x + 5e^{-x}| + C$, we get $p = -1/8$ and $q = 7/8$.

If $\int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx = px + q \cdot \log |4e^x + 5e^{-x}| + C$, then

Similar Questions

Find the following integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$

The value of $\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to

The value of $\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ is

If $\int \frac{d x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(f(x))}{\sqrt{5}}+c$,(where $c$ is the constant of integration),then $f(\pi / 4)$ has the value:

$\int \frac{1}{x^m \sqrt[m]{x^m+1}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module