If int {frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}dx | vedclass

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

Question

(C) Given integral is $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} dx = \int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx$. Using the formula $\int {

Vedclass · Accepted Answer

$\log\, a - \log\, b - \log\, c + 2$

Vedclass · Answer

(C) Given integral is $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} dx = \int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx$. Using the formula $\int {m^x} dx = \frac{{m^x}}{{\ln m}} + C$,where $m = \frac{{a{e^2}}}{{bc}}$,we get: $\int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx = \frac{{{{\left( {\frac{{a{e^2}}}{{bc}}} \right)}^x}}}{{\ln \left( {\frac{{a{e^2}}}{{bc}}} \right)}} + l$. Comparing this with the given form $\frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right) + l$,we identify $k = \ln \left( {\frac{{a{e^2}}}{{bc}}} \right)$. Using logarithmic properties,$k = \ln a + \ln(e^2) - \ln b - \ln c = \ln a + 2 - \ln b - \ln c$. Thus,$k = \ln a - \ln b - \ln c + 2$.

If $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}dx = \frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right)} + l$,then $k =$

Similar Questions

If $\int \frac{e^x-1}{e^x+1} dx = f(x) + c$, then $f(x)$ is equal to

$\int \frac{d x}{\cos (x+4) \cos (x+2)}$ is equal to

Integrate the function: $\sqrt{1-4x^2}$

$\int \sec^2 x \csc^2 x \, dx = $ . . . . . . $+ C$

If $\int \frac{x^4+1}{x^2+1} dx = Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E$,then $A+B+C+D=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module