Math, asked by rishichaurasia89, 3 months ago

integrate (x + 5)/(3x ^ 2 - 2x + 1) dx​

Answers

Answered by Muskansolanki555
0

Step-by-step explanation:

The answer is

=

x

2

2

3

4

ln

(

|

x

+

1

|

)

+

5

4

ln

(

|

x

1

|

)

1

4

ln

(

x

2

+

1

)

+

arctan

x

+

C

Explanation:

We need

d

x

x

2

+

1

=

arctan

(

x

)

As the degree of the numerator is greater than the degree of the denominator, perform a long division first.

x

5

+

3

x

2

+

1

x

4

1

=

x

+

3

x

2

+

x

+

1

x

4

1

Perform a decomposition into partial fractions

3

x

2

+

x

+

1

x

4

1

=

3

x

2

+

x

+

1

(

x

2

+

1

)

(

x

+

1

)

(

x

1

)

=

A

x

+

B

x

2

+

1

+

C

x

+

1

+

D

x

1

=

(

A

x

+

B

)

(

x

+

1

)

(

x

1

)

+

C

(

x

2

+

1

)

(

x

1

)

+

D

(

x

2

+

1

)

(

x

+

1

)

(

x

2

+

1

)

(

x

+

1

)

(

x

1

)

The denominators are the same , compare the numerators

3

x

2

+

x

+

1

=

(

A

x

+

B

)

(

x

+

1

)

(

x

1

)

+

C

(

x

2

+

1

)

(

x

1

)

+

D

(

x

2

+

1

)

(

x

+

1

)

Let

x

=

1

,

,

5

=

4

D

,

,

D

=

5

4

Let

x

=

1

,

,

3

=

4

C

,

,

C

=

3

4

Let

x

=

0

,

, 1=-B-C+D

1

=

B

C

+

D

, =>

, B=5/4+3/4-1=1

B

=

5

4

+

3

4

1

=

1

Coefficients of x^3

x

3

0=A+C+D

0

=

A

+

C

+

D

, =>

, A=-C-D=3/4-5/4=-1/2

A

=

C

D

=

3

4

5

4

=

1

2

Therefore,

(x^5+3x^2+1)/(x^4-1)=x+(-1/2x+1)/(x^2+1)+(-3/4)/(x+1)+(5/4)/(x-1)

x

5

+

3

x

2

+

1

x

4

1

=

x

+

1

2

x

+

1

x

2

+

1

+

3

4

x

+

1

+

5

4

x

1

So,

int((x^5+3x^2+1)dx)/(x^4-1)=intxdx+int((-1/2x+1)dx)/(x^2+1)+int(-3/4dx)/(x+1)+int(5/4dx)/(x-1)

(

x

5

+

3

x

2

+

1

)

d

x

x

4

1

=

x

d

x

+

(

1

2

x

+

1

)

d

x

x

2

+

1

+

3

4

d

x

x

+

1

+

5

4

d

x

x

1

intxdx=x^2/2

x

d

x

=

x

2

2

int(-3/4dx)/(x+1)=-3/4ln(|x+1|)

3

4

d

x

x

+

1

=

3

4

ln

(

|

x

+

1

|

)

int(5/4dx)/(x-1)=5/4ln(|x-1|)

5

4

d

x

x

1

=

5

4

ln

(

|

x

1

|

)

int((-1/2x+1)dx)/(x^2+1)=-1/4int(2xdx)/(x^2+1)+intdx/(x^2+1)=-1/4ln(x^2+1)+arctanx

(

1

2

x

+

1

)

d

x

x

2

+

1

=

1

4

2

x

d

x

x

2

+

1

+

d

x

x

2

+

1

=

1

4

ln

(

x

2

+

1

)

+

arctan

x

Finally,

int((x^5+3x^2+1)dx)/(x^4-1)=x^2/2-3/4ln(|x+1|)+5/4ln(|x-1|)-1/4ln(x^2+1)+arctanx+C

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