If int frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x | vedclass

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

Question

(A) Given the integral $I = \int \frac{3-x^2}{1-2x+x^2} e^x dx$.We can rewrite the denominator as $(1-x)^2$.So, $I = \int \frac{3-x^2}{(1-x)^2} e^x dx

Vedclass · Accepted Answer

$\frac{1+x}{1-x}$

Vedclass · Answer

(A) Given the integral $I = \int \frac{3-x^2}{1-2x+x^2} e^x dx$.We can rewrite the denominator as $(1-x)^2$.So, $I = \int \frac{3-x^2}{(1-x)^2} e^x dx$.We manipulate the numerator to match the form $f(x) + f'(x)$:$I = \int \frac{-(x^2-1) + 2}{(1-x)^2} e^x dx = \int \frac{-(x-1)(x+1) + 2}{(x-1)^2} e^x dx$$I = \int \left( \frac{-(x+1)}{x-1} + \frac{2}{(x-1)^2} \right) e^x dx = \int \left( \frac{x+1}{1-x} + \frac{2}{(1-x)^2} \right) e^x dx$.Let $f(x) = \frac{1+x}{1-x}$. Then $f'(x) = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$.Since $\int (f(x) + f'(x)) e^x dx = e^x f(x) + c$, we have $I = e^x \left( \frac{1+x}{1-x} \right) + c$.Comparing this with $e^x f(x) + c$, we get $f(x) = \frac{1+x}{1-x}$.

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$, then $f(x)$ is:

Similar Questions

Evaluate the integral: $\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \left( {x + \sqrt x } \right)dx$

$\int e^x \tan x(1+\tan x) \, dx = $ . . . . . . $+ C$.

$\int {{e^x}[\tan x - \log (\cos x)]\,dx = }$

$\int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx = $

Integrate the function: $e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module