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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}
Posted by Gpuri Talukdar 8 years, 1 month ago
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Naveen Sharma 8 years, 1 month ago
{tex}\int \sin^4x. \cos x dx \\ put \sin x = t \ then \\ \cos x dx = dt \\ = \int t^4dt\\ = {t^5\over 5} + C\\ = {sin^5x\over 5 } + C{/tex}
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