If int frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d | vedclass

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

Question

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$. Multiply numerator and denominator by $e^x$: $I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x

Vedclass · Accepted Answer

$\frac{3}{4}, \frac{-1}{24}$

Vedclass · Answer

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$. Multiply numerator and denominator by $e^x$: $I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x$. We want to express the numerator in terms of the denominator and its derivative. Let $2 e^{2x} + 3 = A_1(3 e^{2x} + 4) + B_1 \frac{d}{dx}(3 e^{2x} + 4)$. $2 e^{2x} + 3 = 3 A_1 e^{2x} + 4 A_1 + 6 B_1 e^{2x}$. Comparing coefficients: $3 A_1 + 6 B_1 = 2$ and $4 A_1 = 3 \implies A_1 = \frac{3}{4}$. Substituting $A_1$: $3(\frac{3}{4}) + 6 B_1 = 2 \implies \frac{9}{4} + 6 B_1 = 2 \implies 6 B_1 = 2 - \frac{9}{4} = -\frac{1}{4} \implies B_1 = -\frac{1}{24}$. Thus,$I = \int \left( \frac{3}{4} - \frac{1}{24} \frac{6 e^{2x}}{3 e^{2x} + 4} \right) d x$. $I = \frac{3}{4} x - \frac{1}{24} \log(3 e^{2x} + 4) + C$. Comparing with $Ax + B \log(3 e^{2x} + 4) + C$,we get $A = \frac{3}{4}$ and $B = -\frac{1}{24}$.

If $\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C$,then values of $A$ and $B$ are respectively (where $C$ is a constant of integration.)

Similar Questions

$\int \frac{\operatorname{cosec}^2 x-2022}{\cos ^{2022} x} d x=f(x)+C \Rightarrow f(\pi / 4)=$

If $\int(3 x+2) \sqrt{2 x^2+3 x+4} d x=f(x) \sqrt{2 x^2+3 x+4}+A \sinh ^{-1}\left(\frac{4 x+3}{\sqrt{23}}\right)+C$, then the ordered pair $(f(1), A)=$

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

$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ is equal to

The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(2-x^{2}) dx}{(2+x^{2}) \sqrt{4+x^{4}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module