How do you evaluate the integral $\int \frac{x^{4} + 1}{x^{2} + 1}$ $int (x^4+1)/(x^2+1)$ ?

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How do you evaluate the integral $\int \frac{x^{4} + 1}{x^{2} + 1}$ $int (x^4+1)/(x^2+1)$ ?

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Answer 1 · 2017-01-08T13:59:23

The answer is $= \frac{x^{3}}{3} - x + 2 arctan x + C$ $=x^3/3-x+2arctanx+C$

Explanation:

Since the degree of the numerator is not less than the degree of the denominator, perform a long division

$a a a a$ $color(white)(aaaa)$ $x^{4}$ $x^4$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $+ 1$ $+1$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $x^{2} + 1$ $x^2+1$

$a a a a$ $color(white)(aaaa)$ $x^{4} + x^{2}$ $x^4+x^2$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ #color(white)(aa)∣# $x^{2} - 1$ $x^2-1$

$a a a a$ $color(white)(aaaa)$ $0 - x^{2}$ $0-x^2$ $a a a a$ $color(white)(aaaa)$ $+ 1$ $+1$

$a a a a a a$ $color(white)(aaaaaa)$ $- x^{2}$ $-x^2$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$

$a a a a a a$ $color(white)(aaaaaa)$ $0$ $0$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $2$ $2$

Therefore,

$\frac{x^{4} + 1}{x^{2} + 1} = x^{2} - 1 + \frac{2}{x^{2} + 1}$ $(x^4+1)/(x^2+1)=x^2-1+2/(x^2+1)$

$\int \frac{(x^{4} + 1) d x}{x^{2} + 1} = \int x^{2} d x - \int 1 d x + 2 \int \frac{d x}{x^{2} + 1}$ $int((x^4+1)dx)/(x^2+1)=intx^2dx-int1dx+2intdx/(x^2+1)$

$= \frac{x^{3}}{3} - x + 2 \int \frac{d x}{x^{2} + 1}$ $=x^3/3-x+2intdx/(x^2+1)$

Let $x = tan θ$ $x=tan theta$ , $\Rightarrow$ $=>$ , $d x = {sec}^{2} θ d θ$ $dx=sec^2theta d theta$

and $x^{2} + 1 = {tan}^{2} θ + 1 = {sec}^{2} θ$ $x^2+1=tan^2 theta+1=sec^2 theta$

Therefore,

$2 \int \frac{d x}{x^{2} + 1} = 2 \int \frac{{sec}^{2} θ d θ}{{sec}^{2} θ} = 2 \int d θ = 2 θ = 2 arctan x$ $2intdx/(x^2+1)=2int(sec^2 theta d theta)/sec^2theta=2intd theta=2 theta=2arctanx$

So,

$\int \frac{(x^{4} + 1) d x}{x^{2} + 1} = \frac{x^{3}}{3} - x + 2 arctan x + C$ $int((x^4+1)dx)/(x^2+1)=x^3/3-x+2arctanx+C$

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Explanation:

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