x squared plus 1 upon x …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Yashpal Choudhary 8 years, 1 month ago
- 1 answers
Test
Related Questions
Posted by Sanjay Kumar 1 year, 5 months ago
- 0 answers
Posted by Xxxxxx Xx 1 year, 4 months ago
- 3 answers
Posted by Karan Kumar Mohanta 1 year, 4 months ago
- 0 answers
Posted by Sanjna Gupta 1 year, 4 months ago
- 4 answers
Posted by Ananya Singh 1 year, 6 months ago
- 0 answers
Posted by Sneha Pandey 1 year, 5 months ago
- 0 answers
Posted by Xxxxxx Xx 1 year, 4 months ago
- 0 answers
Posted by Charu Baid 1 year, 4 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Naveen Sharma 8 years, 1 month ago
{tex}\int {x^2+1\over x^2-5x+6}dx{/tex}
as this is a improper fraction, convert it into proper fraction. we get
{tex}\int {1}. dx + \int {5x-5\over x^2-5x+6}dx{/tex}
{tex}= x + \int {5x-5\over (x-2)(x-3)}dx{/tex}
Using partial fraction, we get
{tex}{5x-5\over (x-2)(x-3)} = {A\over x-2} + {B\over x-3}{/tex}
=> 5x- 5 = Ax- 3A + Bx-2B
=> 5x - 5 = x(A+B) - (3A+2B)
On comparing both sides, we get
A+B = 5 ... (1)
3A+5B = 5 ..... (2)
Multiply (1) by 3, we get
3A+3B = 15 .... (3)
Subtracting (3) from (2), we get
2B = -10
B = -5
Using value of B, we get A = 10
Now,
{tex}= x + \int {5x-5\over (x-2)(x-3)}dx = x + \int {10\over x-2}dx + \int {-5\over x-3}dx{/tex}
= {tex}x + 10 log|x-2| -5log |x-3| +C{/tex}
1Thank You