Question

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

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

Answer 1

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

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

Factors of $1+x+x^2+x^3$$=(x^2+1)(x+1)$

$\frac{1}{(1+x)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$

$1=A(x^2+1)+(Bx+C)(x+1)$

$1=A{x}^2+A+B{x}^2+Bx+Cx+C$

$1=(A+B)x^2+(B+C)x+A+C$

$A+B=0$

$B+C=0$

$A+C=0$

Therefore, $2A=1$

$A=\frac{1}{2}$

$B=\frac{-1}{2}$

$C=\frac{1}{2}$

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

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

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

$=\frac{1}{2}\ln \mid (x+1)\mid -\frac{1}{4}\int \frac{2xdx}{=}{x}^2+1+2\int \frac{dx}{1+x^2}$

$=x^2+1=u$

$=2xdx=u$

$=\frac{1}{2}\ln \mid (x+1)\mid -\frac{1}{4}\int \frac{du}{u}+2{\tan }^{-1}x+C$

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