Let I=∫exe4x+e2x+1dx,J=e3xe4x+e2x+1then thevalue of I – J equals

I−J=∫exe4x+e2x+1dx−∫e3xe4x+e2x+1dx=∫ex1−e2xe4x+e2x+1dxPut ex=t,exdx=dtI−J=∫1−t2t4+t2+1dt=∫1t2−1t2+1t2+1dt=−∫1−1t2t+1t2−1=∫du1−u2u=t+1t=12log⁡1+u1−u+C=12log⁡ex+e−x+1ex+e−x−1+C=12log⁡e2x+ex+1e2x−ex+1+C.

Let I=∫exe4x+e2x+1dx,J=e3xe4x+e2x+1then thevalue of I – J equals

I−J=∫exe4x+e2x+1dx−∫e3xe4x+e2x+1dx=∫ex1−e2xe4x+e2x+1dxPut ex=t,exdx=dtI−J=∫1−t2t4+t2+1dt=∫1t2−1t2+1t2+1dt=−∫1−1t2t+1t2−1=∫du1−u2u=t+1t=12log⁡1+u1−u+C=12log⁡ex+e−x+1ex+e−x−1+C=12log⁡e2x+ex+1e2x−ex+1+C.

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Let $I = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x, J = \frac{e^{3 x}}{e^{4 x} + e^{2 x} + 1}$ then the
value of $I - J$ equals

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$\frac{1}{2} \log (\frac{e^{4 x} - e^{2 x} + 1}{e^{4 x} + e^{2 x} + 1}) + C$

$\frac{1}{2} \log (\frac{e^{2 x} + e^{x} + 1}{e^{2 x} - e^{x} + 1}) + C$

$\frac{1}{2} \log (\frac{e^{2 x} - e^{x} + 1}{e^{2 x} + e^{x} + 1}) + C$

$\frac{1}{2} \log (\frac{e^{4 x} + e^{2 x} + 1}{e^{4 x} - e^{2 x} + 1}) + C$

answer is B.

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Detailed Solution

$I - J = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x - \int \frac{e^{3 x}}{e^{4 x} + e^{2 x} + 1} d x$

$= \int \frac{e^{x} (1 - e^{2 x})}{e^{4 x} + e^{2 x} + 1} d x$

Put $e^{x} = t, e^{x} d x = d t$

$\begin{array}{l} I - J = \int \frac{(1 - t^{2})}{t^{4} + t^{2} + 1} d t = \int \frac{\frac{1}{t^{2}} - 1}{t^{2} + \frac{1}{t^{2}} + 1} d t \\ = - \int \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 1} = \int \frac{d u}{1 - u^{2}} \\ u = t + \frac{1}{t} \end{array}$

$\begin{aligned} = \frac{1}{2} \log |\frac{1 + u}{1 - u}| + C \\ = \frac{1}{2} \log |\frac{e^{x} + e^{- x} + 1}{e^{x} + e^{- x} - 1}| + C \\ = \frac{1}{2} \log |\frac{e^{2 x} + e^{x} + 1}{e^{2 x} - e^{x} + 1}| + C . \end{aligned}$