Question

$\int \frac{dx}{1+x+x^2+x^3}=$

• A. $log\sqrt{1+x}-\frac{1}{2}log\sqrt{1+x^2}+\frac{1}{2}{\tan }^{-1}x+c$
• B. $log\sqrt{1+x}-log\sqrt{1+x^2}+{\tan }^{-1}x+c$
• C. $log\sqrt{1+x^2}-log\sqrt{1+x}+\frac{1}{2}{\tan }^{-1}x+c$
• D. $log\sqrt{1+x}+{\tan }^{-1}x+log\sqrt{1+x^2}+c$

Answer 1

The correct option is A $log\sqrt{1+x}-\frac{1}{2}log\sqrt{1+x^2}+\frac{1}{2}{\tan }^{-1}x+c$

$\int \frac{1}{x^3+x^2+x+1}dx$

$=\int \frac{1}{(x+1)(x^2+1)}dx$

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

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

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

Substitute $u=x^2+1$

$dx=\frac{1}{2x}dx$

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

$=\frac{1}{2}ln(x+1)-\frac{1}{2}\left[\int \frac{x}{x^2+1}dx-\int \frac{1}{x^2+1}dx\right]$

$=\frac{1}{2}ln(x+1)-\frac{1}{2}\left[\frac{1}{2}\int \frac{1}{x}dx-\int \frac{1}{x^2+1}dx\right]$

$=\frac{1}{2}ln(x+1)-\frac{1}{4}ln(x^2+1)-\frac{1}{2}{\tan }^{-1}x$

$=\frac{ln(x+1)}{2}-\frac{ln(x^2+1)}{4}-\frac{ta{n}^{-1}x}{2}$.