Practicing Success
\(\int\limits_1^2\frac{dx}{x\left(x^4+1\right)} = ? \) |
\(\log {\left(\frac{32}{17}\right)}\) \(\log {\left(\frac{16}{17}\right)}\) \( \frac{1}{4} \log {\left(\frac{16}{17}\right)}\) \( \frac{1}{4} \log {\left(\frac{32}{17}\right)}\) |
\( \frac{1}{4} \log {\left(\frac{32}{17}\right)}\) |
\(\int\limits_1^2\frac{dx}{x\left(x^4+1\right)}=\int\limits_1^2\frac{dx}{x^5(1+x^{-4})}\) Let $t= 1+x^{-4}$ $ x=1 \rightarrow , t=2 $ $x=2 , \rightarrow t=\frac{17}{16}$
$-4x^{-5}dx=dt$ $\int\limits_2^{17/16}\frac{-1}{4t}dt=-\frac{1}{4}[ln\, t]_2^{17/16}⇒-\frac{1}{4}[ln\frac{17}{16}-ln2]$ $=\frac{1}{4}ln[\frac{32}{17}]⇒\frac{1}{4}log[\frac{32}{17}]$ So, option D is correct. |