Question

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

• A. $\frac{x^3}{3}+x+\frac{x}{x^2+1}+c$
• B. $\frac{x^5+x^3+x+3}{3(x^2+1)}+c$
• C. $\frac{x^5+4x^3+3x+3}{3(x^2+1)}+c$
• D. None of these

Answer 1

Answer: (D)

None of these

$I=\int \frac{(x-1{)}^2}{x^4+2x^2+1}dx=\int \frac{{(x-1)}^2}{{(x^2+1)}^2}dx$
$=\int \frac{(x^2+1-2x)}{{(x^2+1)}^2}dx$
$=\int (\frac{1}{x^2+1}-\frac{2x}{{(x^2+1)}^2})dx$
Put in second term, $x^2+1=t⟹2xdx=dt$
$={\tan }^{-1}x-\int \frac{1}{t^2}dt+c$
$={\tan }^{-1}x+\frac{1}{t}+c$
$={\tan }^{-1}x+\frac{1}{x^2+1}+c$
$=\frac{x^2{\tan }^{-1}x+{\tan }^{-1}x+1}{x^2+1}+c$