If I=int frac{e^x}{e^{4 x}+e^{2 x}+1} ,d x an | vedclass

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

Question

(C) Given $J-I = \int \left( \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} - \frac{e^x}{e^{4x} + e^{2x} + 1} \right) dx$.Multiply the numerator and denominator

Vedclass · Accepted Answer

$\frac{1}{2} \log \left|\left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)\right|+c$

Vedclass · Answer

(C) Given $J-I = \int \left( \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} - \frac{e^x}{e^{4x} + e^{2x} + 1} \right) dx$.Multiply the numerator and denominator of the first term by $e^{4x}$: $\frac{e^{-x} \cdot e^{4x}}{e^{-4x} \cdot e^{4x} + e^{-2x} \cdot e^{4x} + 1 \cdot e^{4x}} = \frac{e^{3x}}{1 + e^{2x} + e^{4x}}$.Thus, $J-I = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1} dx = \int \frac{(e^{2x} - 1)e^x}{e^{4x} + e^{2x} + 1} dx$.Let $e^x = t$, then $e^x dx = dt$.The integral becomes $\int \frac{t^2 - 1}{t^4 + t^2 + 1} dt = \int \frac{1 - 1/t^2}{(t^2 + 1 + 1/t^2)} dt = \int \frac{1 - 1/t^2}{(t + 1/t)^2 - 1} dt$.Let $u = t + 1/t$, then $du = (1 - 1/t^2) dt$.The integral is $\int \frac{du}{u^2 - 1} = \frac{1}{2} \log \left| \frac{u-1}{u+1} \right| + c$.Substituting $u = t + 1/t = e^x + e^{-x}$, we get $\frac{1}{2} \log \left| \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right| + c = \frac{1}{2} \log \left| \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right| + c$.

If $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} \,d x$ and $J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} \,d x$ then for any arbitrary constant $c$, the value of $J-I$ equals

Similar Questions

Let $f(x) = x$, $f_1(x) = f(\log x)$, $f_2(x) = f_1(\log x)$, $f_3(x) = f_2(\log x) \dots$ and so on. Then $\int \frac{1}{f(x) f_1(x) f_2(x) \dots f_{2026}(x)} dx = \dots$

$\int \sqrt{\frac{1-x}{1+x}} \, dx = $

If $\int f(x) dx = g(x) + c$, then $\int f^{-1}(x) dx =$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module