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

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

Question

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx$. We express the numerator as $4e^{2x} + 6 = m(18e^{2x}) + n(9e^{2x} - 4)$. Comparing coefficients: $18m + 9n = 4$ and $-4n = 6 \implies n = -\frac{3}{2}$. $18m + 9(-\frac{3}{2}) = 4 \implies 18m = 4 + \frac{27}{2} = \frac{35}{2} \implies m = \frac{35}{36}$. Thus,$I = \int \frac{\frac{35}{36}(18e^{2x}) - \frac{3}{2}(9e^{2x} - 4)}{9e^{2x} - 4} dx = \frac{35}{36} \int \frac{18e^{2x}}{9e^{2x} - 4} dx - \frac{3}{2} \int dx$. $I = \frac{35}{36} \log |9e^{2x} - 4| - \frac{3}{2}x + C$. Comparing with $Ax + B \log |9e^{2x} - 4| + C$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$. Since none of the options match these values,the correct answer is $(d)$.

If $\int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx = Ax + B \log |9e^{2x} - 4| + C$,then $A, B$ and $C$ are

Similar Questions

$\int \frac{2 x+2}{\sqrt{x^2-4 x-5}} d x$ is equal to

If $f(x) = \int {\left( {\frac{{{x^2} + {{\sin }^2}x}}{{1 + {x^2}}}} \right)} {\sec ^2}x\,dx$ and $f(0) = 0,$ then $f(1)$ equals

If $\int \frac{dx}{\sin^3 x + \cos^3 x} = A \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right| + B \tan^{-1}(t) + c$,then $\left(\frac{B}{A}, t\right) =$

Given that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. If $\int \frac{1}{x^4+3x^2+1} dx = a \cdot \tan^{-1}\left(\frac{b(x^2-1)}{x}\right) + c \cdot \tan^{-1}\left(\frac{d(x^2+1)}{x}\right) + k$, where $k$ is a constant of integration, then $5(c+d+ab) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module