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Answer:$\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c$

Hint: To solve this integral, use special integral formula.

Given: $\int \frac{x^{4}+1}{x^{2}+1} d x$

Solution:

Let

\begin{aligned} &I=\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{x^{4}+1+2 x^{2}-2 x^{2}}{x^{2}+1} d x \\\\ &=\int \frac{\left\{\left(x^{2}\right)^{2}+1+2 x^{2}\right\}-2 x^{2}}{x^{2}+1} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \end{aligned}

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

\begin{aligned} &=\int\left\{\left(x^{2}+1\right)-\frac{\left[2 x^{2}+2-2\right]}{x^{2}+1}\right\} d x \\\\ &=\int\left\{\left(x^{2}+1\right)-\frac{\left[2\left(x^{2}+1\right)-2\right]}{x^{2}+1}\right\} d x \\\\ &=\int\left\{\left(x^{2}+1\right)-\frac{2\left(x^{2}+1\right)}{x^{2}+1}+\frac{2}{x^{2}+1}\right\} d x \end{aligned}

\begin{aligned} &=\int\left\{\left(x^{2}+1\right)-2+\frac{2}{x^{2}+1}\right\} d x \\\\ &=\int\left(x^{2}+1\right) d x-2 \int 1 d x+2 \int \frac{1}{x^{2}+1} d x \\ \\&=\int x^{2} d x+\int 1 d x-2 \int 1 d x+2 \int \frac{1}{x^{2}+1} d x \end{aligned}

$=\frac{x^{2+1}}{2+1}+x-2 x+2 \cdot \frac{1}{1} \tan ^{-1}\left(\frac{x}{1}\right)+c \quad \quad \quad \quad \quad \quad\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, \int 1 d x=x+c \\\\ \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \end{array}\right]$
$=\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c$

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