Integrate the function $\frac{3}{x-2}+\frac{2x+3}{x^2+2x+4}+\frac{-\left(6x+12\right)}{x^3-8}$

Step-by-step Solution

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Final answer to the problem

$\ln\left|x-2\right|+\frac{1}{\sqrt{3}}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+4\ln\left|\sqrt{\left(x+1\right)^2+3}\right|+C_1$
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Step-by-step Solution

How should I solve this problem?

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

$\int\left(\frac{3}{x-2}+\frac{2x+3}{x^2+2x+4}+\frac{-\left(6x+12\right)}{x^3-8}\right)dx$

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$\int\left(\frac{3}{x-2}+\frac{2x+3}{x^2+2x+4}+\frac{-\left(6x+12\right)}{x^3-8}\right)dx$

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Unlock the first 3 steps of this solution

Learn how to solve problems step by step online. Integrate the function 3/(x-2)+(2x+3)/(x^2+2x+4)(-(6x+12))/(x^3-8). Find the integral. Simplify the expression. The integral \int\frac{3}{x-2}dx results in: 3\ln\left(x-2\right). The integral \int\frac{-6x-12}{x^3-8}dx results in: -\int\frac{6x+12}{x^3-8}dx.

Final answer to the problem

$\ln\left|x-2\right|+\frac{1}{\sqrt{3}}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+4\ln\left|\sqrt{\left(x+1\right)^2+3}\right|+C_1$

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Function Plot

Plotting: $\ln\left(x-2\right)+\frac{1}{\sqrt{3}}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+4\ln\left(\sqrt{\left(x+1\right)^2+3}\right)+C_1$

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a
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g
m
n
u
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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

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