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Integrate the function $\frac{x^5-x^4+x^2-2}{x^2+1}$

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Final answer to the problem

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

How should I solve this problem?

  • Integrate by parts
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
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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

$\int x^{3}dx+\int-x^{2}dx+\int-xdx+\int2dx+\int\frac{x-4}{x^2+1}dx$
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5

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

$\frac{x^{4}}{4}$
6

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

$\frac{-x^{3}}{3}$
7

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

$-\frac{1}{2}x^2$
8

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

$2x$
9

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

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

Gather the results of all integrals

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

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

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

Final answer to the problem

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

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Function Plot

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

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0
a
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d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Equations

In mathematics, an equation is a statement of an equality containing one or more variables. Solving the equation consists of determining which values of the variables make the equality true. In this situation, variables are also known as unknowns and the values which satisfy the equality are known as solutions.

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