Question

$\int \frac{x^2(1-\ell nx)}{(\ell n^{4}x-x^4)}dx$ equals

• A. $\frac{1}{4}\ell n\left(\frac{x}{\ell nx}\right)-\frac{1}{4}\ell n(\ell n^{2}x-x^2)+c$
• B. $\frac{1}{4}\ell n\left(\frac{\ell nx-x}{\ell nx+x}\right)-\frac{1}{2}ta{n}^{-1}\left(\frac{\ell nx}{x}\right)+c$
• C. $\frac{1}{4}\ell n\left(\frac{\ell nx+x}{\ell nx-x}\right)+\frac{1}{2}ta{n}^{-1}\left(\frac{\ell nx}{x}\right)+c$
• D. $\frac{1}{4}\left(\ell n\right(\frac{\ell nx-x}{\ell nx-x})-ta{n}^{-1}(\frac{\ell nx}{x}\left)\right)+c$

Answer 1

The correct option is B $\frac{1}{4}\ell n\left(\frac{\ell nx-x}{\ell nx+x}\right)-\frac{1}{2}ta{n}^{-1}\left(\frac{\ell nx}{x}\right)+c$

$I=\int \frac{x^2(1-\ell nx)dx}{(\ell n^{4}x-x^4)}=\frac{x^2(1-\ell nx)dx}{x^4\left(\right(\frac{\ell nx}{x}{)}^4)-1}=\int \frac{(1-\ell nx)dx}{x^2\left(\right(\frac{\ell nx}{x}{)}^4)-1}Put\frac{\ell nx}{x}=t\to \left(\frac{1-\ell nx}{x^2}\right)dx=dtI=\int \frac{dt}{(t^4-1)}=\int \frac{dt}{(t^2+1)(t^2-1)}=\frac{1}{2}\int \frac{(t^2+1)-(t^2-1)dt}{(t^2+1)(t^2-1)}I=\frac{1}{2}\int \frac{dt}{t^2-1}-\frac{1}{2}\int \frac{dt}{t^2+1}=\frac{1}{2}(\frac{1}{2}\ell n\int \frac{t-1}{t+1})-ta{n}^{-1}t=\frac{1}{4}\ell n\left(\frac{\ell nx-x}{\ell nx+x}\right)-\frac{1}{2}ta{n}^{-1}\left(\frac{\ell nx}{x}\right)+c$