How do you find #int 10/((x-1)(x^2+9)) dx# using partial fractions?

Answer 1

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

Since the denominator consists of 1 linear factor and 1 irreducible quadratic factor, we may use the method of partial fractions to rewrite it as :

#10/((x-1)(x^2+9))=A/(x-1)+(Bx+C)/(x^2+9)#
#=(A(x^2+9)+(Bx+C)(x-1))/((x-1)(x^2+9))#
#therefore 10 = Ax^2+9A+Bx^2-Bx+Cx-C#
#= (A+B)x^2-(B-C)x+(9A-C)#

Now comparing terms we get that

#A+B=0# #C-B=0# #9A-C=10#

Solving this linear system of equations yields :

#A=1, B=-1, C=-1#

Therefore the original integral may be written and solved as

#int10/((x-1)(x^2+9))dx=int1/(x-1)dx-int(x)/(x^2+9)dx-int1/(x^2+9)dx#
#=ln|x-1|-1/2ln|x^2+9|+1/3tan^(-1)(x/3)+C#
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Answer 2

To find the integral of ( \frac{10}{(x-1)(x^2+9)} ) using partial fractions, we first decompose the fraction into partial fractions. We express ( \frac{10}{(x-1)(x^2+9)} ) as ( \frac{A}{x-1} + \frac{Bx + C}{x^2+9} ). Then, we find the values of A, B, and C by equating numerators.

Multiplying both sides by the denominator ( (x-1)(x^2+9) ), we get: [ 10 = A(x^2 + 9) + (Bx + C)(x - 1) ]

Expanding and collecting like terms, we get: [ 10 = Ax^2 + 9A + Bx^2 - Bx + Cx - C ]

Matching coefficients for like terms, we have: [ (A + B)x^2 - Bx + Cx + (9A - C) = 10 ]

This equation implies that: [ A + B = 0 ] [ -B + C = 0 ] [ 9A - C = 10 ]

Solving this system of equations, we find: [ A = \frac{5}{3} ] [ B = -\frac{5}{3} ] [ C = -\frac{5}{3} ]

Now, we rewrite the original integral using the partial fraction decomposition: [ \int \frac{10}{(x-1)(x^2+9)} ,dx = \int \left( \frac{5}{x-1} - \frac{5x + 5}{x^2 + 9} \right) ,dx ]

We can now integrate each term separately: [ \int \frac{5}{x-1} ,dx = 5\ln|x-1| + C_1 ] [ \int \frac{-5x - 5}{x^2 + 9} ,dx = -\frac{5}{2}\ln|x^2 + 9| + C_2 ]

Where ( C_1 ) and ( C_2 ) are constants of integration.

Therefore, the integral of ( \frac{10}{(x-1)(x^2+9)} ) using partial fractions is: [ 5\ln|x-1| - \frac{5}{2}\ln|x^2 + 9| + C ]

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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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