Q84 / 160 · TS EAMCET 2017 · Mathematics
$\int \frac{d x}{x\left(x^4+1\right)}=$
- A$\frac{1}{4} \log \left(\frac{x^4+1}{x^4}\right)+C$
- Correct answerB$\frac{1}{4} \log \left(\frac{x^4}{x^4+1}\right)+C$
- C$\frac{1}{4} \log \left(x^4+1\right)+C$
- D$\frac{1}{4} \log \left(\frac{x^4}{x^4+2}\right)+C$
Solution
$$
\text { } \begin{aligned}
\int \frac{d x}{x\left(x^4+1\right)} & =\int \frac{x^4+1-x^4}{x\left(x^4+1\right)} d x \\
& =\int \frac{x^4+1}{x\left(x^4+1\right)} d x-\int \frac{x^4}{x\left(x^4+1\right)} d x \\
& =\int \frac{1}{x} d x-\int \frac{x^3}{x^4+1} d x \\
& =\log |x|-\int \frac{x^3}{x^4+1} d x+C
\end{aligned}
$$
Let $x^4+1=t \quad \Rightarrow \quad x^3 d x=d t$
$$
\Rightarrow \quad x^3 d x=\frac{1}{4}
$$
$$
\begin{aligned}
\therefore \int \frac{d x}{4\left(x^4+1\right)} & =\log |x|-\frac{1}{4} \int \frac{d t}{t}+C \\
& =\log |x|-\frac{1}{4} \log |t|+C \\
& =\log |x|-\frac{1}{4} \log \left|x^4+1\right|+C \\
& =\frac{1}{4} \log \left|x^4\right|-\frac{1}{4} \log \left|x^4+1\right|+C \\
& =\frac{1}{4}\left[\log \left|x^4\right|-\log \left|x^4+1\right|\right]+C \\
& =\frac{1}{4} \log \left|\frac{x^4}{x^4+1}\right|+C
\end{aligned}
$$
Hence, the correct answer is option: 'B'.