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$\int \frac{d x}{1+x+x^2+x^3}=$
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Verified Answer
The correct answer is:
$\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+\frac{1}{2} \tan ^{-1} x+c$
$\begin{aligned} & \int \frac{d x}{1+x+x^2+x^3}=\int \frac{d x}{(1+x)\left(1+x^2\right)} \\
& =\frac{1}{2} \int \frac{1}{1+x^2} d x+\frac{1}{2} \int \frac{1}{1+x} d x-\frac{1}{2} \int \frac{x}{1+x^2} d x\end{aligned}$
\(\Rightarrow\) As, assume,
\(\frac{1}{(1+x)\left(1+x^2\right)}=\frac{A}{(1+x)}+\frac{B x+c}{\left(1+x^2\right)}\)
By solving it,
\(\begin{aligned}
& A=\frac{1}{2}, B=\frac{-1}{2}, \\
& c=\frac{1}{2} \\
& =\frac{1}{2} \tan ^{-1} x+\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+c
\end{aligned}\)
& =\frac{1}{2} \int \frac{1}{1+x^2} d x+\frac{1}{2} \int \frac{1}{1+x} d x-\frac{1}{2} \int \frac{x}{1+x^2} d x\end{aligned}$
\(\Rightarrow\) As, assume,
\(\frac{1}{(1+x)\left(1+x^2\right)}=\frac{A}{(1+x)}+\frac{B x+c}{\left(1+x^2\right)}\)
By solving it,
\(\begin{aligned}
& A=\frac{1}{2}, B=\frac{-1}{2}, \\
& c=\frac{1}{2} \\
& =\frac{1}{2} \tan ^{-1} x+\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+c
\end{aligned}\)
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