Question

$\int \frac{x^2-1dx}{x\sqrt{x^4+4x^3-6x^2+4x+1}}$ equals

• A. $\ln |x+\frac{1}{x}+\sqrt{(x+\frac{1}{x}+2{)}^2-12}|+C$
• B. $(x+\frac{1}{x})+\ln \left|\right\{(x+\frac{1}{x}+2{)}^2-12\}|+C$
• C. $\ln |x+\frac{1}{x}+2+\sqrt{(x+\frac{1}{x}+2{)}^2-12}|+C$
• D. $(x-\frac{1}{x}{)}^2+\ln \left|\right\{(x-\frac{1}{x}+12{)}^2-12\}|+C$

Answer 1

The correct option is C

$\ln |x+\frac{1}{x}+2+\sqrt{(x+\frac{1}{x}+2{)}^2-12}|+C$

$I=\int \frac{x^2-1dx}{x^2\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}=\int \frac{(1-\frac{1}{x^2})dx}{\sqrt{(x^2+\frac{1}{x^2})+4(x+\frac{1}{x})-6}}$

$=\int \frac{x^2-1dx}{\sqrt{(x+\frac{1}{x}{)}^2+4(x+\frac{1}{x})-8}}$

Put $(x+\frac{1}{x})=t,$ we get $(1-\frac{1}{x^2})dx=dt$

$\therefore I=\int \frac{dt}{\sqrt{t^2+4t-8}}=\int \frac{dt}{\sqrt{(t+2{)}^2-(\sqrt{12}{)}^2}}$

$=\ln \left|\right(t+2)+\sqrt{(t+2{)}^2-(\sqrt{12}{)}^2}|+c$

$=\ln |x+\frac{1}{x}+2+\sqrt{(x+\frac{1}{x}+2{)}^2-12}|+c$