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Evaluate the following integral:

Question:

Evaluate the following integral:

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

Solution:

$I=\int \frac{1}{1+x+x^{2}+x^{3}}=\int \frac{d x}{\left(x^{2}+1\right)(x+1)}$

$\frac{1}{\left(x^{2}+1\right)(x+1)}=\frac{A x+B}{x^{2}+1}+\frac{C}{x+1}$

$1=(A x+B)(x+1)+C\left(x^{2}+1\right)$

Equating constants

$1=B+C$

Equating coefficients of $x$

$0=A+C$

Solving, we get

$\mathrm{A}=-\frac{1}{2} \mathrm{~B}=\frac{1}{2} \mathrm{C}=\frac{1}{2}$

Thus

$I=-\frac{1}{2} \int \frac{x d x}{x^{2}+1}+\frac{1}{2} \int \frac{d x}{x^{2}+1}+\frac{1}{2} \int \frac{d x}{x+1}$

$I=-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x+1|+C$

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