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# Find the integral of $\frac{x^5-x^4+x^2-2}{x^2+1}$

## Step-by-step Solution

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###  Solution

$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$
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##  Step-by-step Solution 

Problem to solve:

$\int\frac{x^5-x^4+x^2-2}{x^2+1}dx$

Specify the solving method

1

Find the integral

$\int\frac{x^5-x^4+x^2-2}{x^2+1}dx$
2

Divide $x^5-x^4+x^2-2$ by $x^2+1$

$\begin{array}{l}\phantom{\phantom{;}x^{2}+1;}{\phantom{;}x^{3}-x^{2}-x\phantom{;}+2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+1\overline{\smash{)}\phantom{;}x^{5}-x^{4}\phantom{-;x^n}+x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x^{2}+1;}\underline{-x^{5}\phantom{-;x^n}-x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{5}-x^{3};}-x^{4}-x^{3}+x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n;}\underline{\phantom{;}x^{4}\phantom{-;x^n}+x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{;\phantom{;}x^{4}+x^{2}-;x^n;}-x^{3}+2x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n-;x^n;}\underline{\phantom{;}x^{3}\phantom{-;x^n}+x\phantom{;}\phantom{-;x^n}}\\\phantom{;;\phantom{;}x^{3}+x\phantom{;}-;x^n-;x^n;}\phantom{;}2x^{2}+x\phantom{;}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n-;x^n-;x^n;}\underline{-2x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}}\\\phantom{;;;-2x^{2}-2\phantom{;}\phantom{;}-;x^n-;x^n-;x^n;}\phantom{;}x\phantom{;}-4\phantom{;}\phantom{;}\\\end{array}$
3

Resulting polynomial

$\int\left(x^{3}-x^{2}-x+2+\frac{x-4}{x^2+1}\right)dx$
4

Expand the integral $\int\left(x^{3}-x^{2}-x+2+\frac{x-4}{x^2+1}\right)dx$ into $5$ integrals using the sum rule for integrals, to then solve each integral separately

$\int x^{3}dx+\int-x^{2}dx+\int-xdx+\int2dx+\int\frac{x-4}{x^2+1}dx$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $3$

$\frac{x^{4}}{4}$
5

The integral $\int x^{3}dx$ results in: $\frac{x^{4}}{4}$

$\frac{x^{4}}{4}$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int x^{2}dx$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$\frac{-x^{3}}{3}$
6

The integral $\int-x^{2}dx$ results in: $\frac{-x^{3}}{3}$

$\frac{-x^{3}}{3}$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int xdx$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$-\frac{1}{2}x^2$
7

The integral $\int-xdx$ results in: $-\frac{1}{2}x^2$

$-\frac{1}{2}x^2$

The integral of a constant is equal to the constant times the integral's variable

$2x$
8

The integral $\int2dx$ results in: $2x$

$2x$

Expand the fraction $\frac{x-4}{x^2+1}$ into $2$ simpler fractions with common denominator $x^2+1$

$\int\left(\frac{x}{x^2+1}+\frac{-4}{x^2+1}\right)dx$

Expand the integral $\int\left(\frac{x}{x^2+1}+\frac{-4}{x^2+1}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{x}{x^2+1}dx+\int\frac{-4}{x^2+1}dx$

The integral of a function times a constant ($-4$) is equal to the constant times the integral of the function

$\int\frac{x}{x^2+1}dx-4\int\frac{1}{1+x^2}dx$

Solve the integral by applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

$\int\frac{x}{x^2+1}dx-4\cdot \left(\frac{1}{\sqrt{1}}\right)\arctan\left(\frac{x}{\sqrt{1}}\right)$

Simplify the expression inside the integral

$\int\frac{x}{x^2+1}dx-4\arctan\left(x\right)$

We can solve the integral $\int\frac{x}{x^2+1}dx$ by applying integration method of trigonometric substitution using the substitution

$x=\tan\left(\theta \right)$

Now, in order to rewrite $d\theta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

$dx=\sec\left(\theta \right)^2d\theta$

Substituting in the original integral, we get

$\int\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{\tan\left(\theta \right)^2+1}d\theta-4\arctan\left(x\right)$

Applying the trigonometric identity: $\tan(x)^2+1=\sec(x)^2$

$\int\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{\sec\left(\theta \right)^2}d\theta-4\arctan\left(x\right)$
Why is tan(x)^2+1 = sec(x)^2 ?

Simplify the fraction $\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{\sec\left(\theta \right)^2}$ by $\sec\left(\theta \right)^2$

$\int\tan\left(\theta \right)d\theta-4\arctan\left(x\right)$

The integral of the tangent function is given by the following formula, $\displaystyle\int\tan(x)dx=-\ln(\cos(x))$

$-\ln\left(\cos\left(\theta \right)\right)-4\arctan\left(x\right)$

Express the variable $\theta$ in terms of the original variable $x$

$\frac{1}{2}\ln\left(x^2+1\right)-4\arctan\left(x\right)$
9

The integral $\int\frac{x-4}{x^2+1}dx$ results in: $\frac{1}{2}\ln\left(x^2+1\right)-4\arctan\left(x\right)$

$\frac{1}{2}\ln\left(x^2+1\right)-4\arctan\left(x\right)$
10

Gather the results of all integrals

$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)$
11

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$

$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$

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ln
log
log
lim
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Dx
|◻|
θ
=
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<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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