What is the integral of #(x^2 +2x-1)/(x^2+9)#?

Answer 1

#=x+ln(x^2+9)-10/3tan^-1(x/3)+C#

First of all split the fraction up like so:

#(x^2+2x-1)/(x^2+9) =x^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)#

We can now integrate each fraction one by one, i.e:

#int(x^2+2x-1)/(x^2+9)dx =intx^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx#

We will have to do a bit of rearranging of the first fraction (add and subtract 9):

#intx^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx=#
#=int(x^2+9-9)/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx#

This can now be rearranged like so:

#=int(x^2+9)/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)-9/(x^2+9)dx#
#=int1+(2x)/(x^2+9) -10/(x^2+9)dx#
-The first term obviously just integrates to #x#.
-For the second term we should apply: #int(f'(x))/f(x)=ln(f(x))+C#
#int (2x)/(x^2+9)dx=ln(x^2+9)+C#

-And for the third term:

#int10/(x^2+9)dx#
Use the substitution #x=3tan(u)# #-> dx = 3 sec^2(u)du#
We also need the trig identity: #tan^2(x)+1=sec^2(x)# Putting the substitution in:
#10int(3sec^2(u))/(9tan^2(u)+9)du=10/3intsec^2(u)/sec^2(u)du#
#=10/3intdu=10/3u+C#

Reverse the substitution and we get:

#10/3tan^-1(x/3)+C#

So returning to our original integral if apply what have found we get that:

#int(x^2+2x-1)/(x^2+9)dx=int1+(2x)/(x^2+9) -10/(x^2+9)dx#
#=x+ln(x^2+9)-10/3tan^-1(x/3)+C#
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Answer 2

To find the integral of ((x^2 + 2x - 1)/(x^2 + 9)), you can use the method of partial fraction decomposition. After decomposing the fraction into simpler fractions, you can integrate each term individually.

The integral of ((x^2 + 2x - 1)/(x^2 + 9)) can be expressed as: [ \int \frac{x^2 + 2x - 1}{x^2 + 9} ,dx = \int \left(1 - \frac{9}{x^2 + 9}\right) ,dx ]

Then, integrating each term separately: [ \int 1 ,dx = x ] [ \int \frac{9}{x^2 + 9} ,dx = 3 \arctan\left(\frac{x}{3}\right) ]

Therefore, the integral of ((x^2 + 2x - 1)/(x^2 + 9)) is: [ x - 3 \arctan\left(\frac{x}{3}\right) + C ] where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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