Mathematics | TS EAMCET 2017 | Discussion Page

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

$\frac{1}{4}\log\left(\frac{x^{4}+1}{x^{4}}\right)+C$

$\frac{1}{4}\log\left(\frac{x^{4}}{x^{4}+1}\right)+C$

$\frac{1}{4} \log (x^{4}+1)+C$

$\frac{1}{4}\log\left(\frac{x^{4}}{x^{4}+2}\right)+C$

Option B

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

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

$=\int\frac{1}{x}dx-\int\frac{x^{3}}{x^{4}+1}dx$

$=\log|x| dx-\int\frac{x^{3}}{x^{4}+1}dx+C$

Let $x^{4}+1=t \Rightarrow x^{3}dx=dt$

$\Rightarrow$ $x^{3} dx=\frac{1}{4}$

$\therefore\int \frac{dx}{4(x^{4}+1)}=\log |x|-\frac{1}{4}\int \frac{dt}{t}+C$

= $\log |x|-\frac{1}{4} \log |t|+C$

= $\log|x|- \frac{1}{4} \log |x^{4}+1|+C$

= $\frac{1}{4} \log |x^{4}|-\frac{1}{4} \log |x^{4}+1|+C$

$=\frac{1}{4}[ \log |x^{4}|-\log |x^{4}+1|+C$

= $\frac{1}{4}\log\left(\frac{x^{4}}{x^{4}+1}\right)+C$

Discussion Forum