Question

If $\int \frac{1+e^{7\ln x}-e^{3\ln x}}{e^{2\ln x}-1}dx=a\ln \left|\frac{x+b}{x-b}\right|+g\left(x\right)+C,g\left(0\right)=0.$ Then which of the following option(s) is/are correct
(where $a,b$ are fixed constants and $C$ is constant of integration)

• A. $2a-b=0$
• B. $2a+b=0$
• C. $g^{\prime }\left(1\right)=2$
• D. $g\left(1\right)=2$

Answer 1

Answer: (B), (C)

$g^{\prime }\left(1\right)=2$

Let $I=\int \frac{1+e^{7\ln x}-e^{3\ln x}}{e^{2\ln x}-1}dx$
$\Rightarrow I=\int \frac{1+x^7-x^3}{x^2-1}dx$
$\Rightarrow I=\int \frac{1}{x^2-1}dx+\int \frac{x^3(x^4-1)}{x^2-1}dx$
$\Rightarrow I=\int \frac{1}{x^2-1}dx+\int \frac{x^3(x^2-1)(x^2+1)}{x^2-1}dx$
$\Rightarrow I=\int \frac{1}{x^2-1}dx+\int x^{5}dx+\int x^{3}dx$
$\Rightarrow I=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+\frac{x^6}{6}+\frac{x^4}{4}+C$
$\Rightarrow I=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+\frac{x^4(2x^2+3)}{12}+C$
$\therefore a=\frac{1}{2},b=-1,g\left(x\right)=\frac{x^4(2x^2+3)}{12}$
$\Rightarrow g\left(1\right)=\frac{5}{12}$ and
$g^{\prime }\left(x\right)=\frac{12x^5+12x^3}{12}\Rightarrow g^{\prime }\left(1\right)=2$