Question

If $\int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}dx=\int \{1+\frac{f\left(x\right)}{(x^2+3)(x^2-5)}\}dx$
$x+A{\tan }^{-1}\left(\frac{x}{A^{\prime }}\right)+B\log \left(\frac{x-l}{x+m}\right)+K$ then which of the following is correct

• A. $A=\frac{1}{4\sqrt{3}},B=\frac{27}{8\sqrt{5}},K\in R$
• B. $f\left(x\right)=7x^2+19,A^{\prime }=\sqrt{3},K\in R$
• C. $l=m=\sqrt{5},L=1,K\in R$
• D. All of these

Answer 1

The correct option is D All of these

$=\int 1+\frac{7x^2+19}{(x^2+3)(x^2-5)}dx=\int (1+\frac{1}{4(x^2+3)}+\frac{27}{4}\frac{1}{x^2-5})dx$
$\therefore \int \frac{{(x}^2+1\left)\right(x^2+4)}{(x^2+3)(x^2-5)}dx=x+\frac{1}{4\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{27}{8\sqrt{5}}\log \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)+K$
$\therefore \int \frac{{(x}^2+1\left)\right(x^2+4)}{(x^2+3)(x^2-5)}dx=Lx+A{\tan }^{-1}\left(\frac{x}{A^{\prime }}\right)+B\log \left(\frac{x-l}{x+m}\right)+K$
$\therefore L=1,A=\frac{1}{4\sqrt{3}},A^{\prime }=\sqrt{3},B=\frac{27}{8\sqrt{5}},K\in R$
$l=m=\sqrt{5};f\left(x\right)=7x^2+19$