Question

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

Answer 1

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:-   by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1  =  C

step 3:-

substitute A,B,C values in equation (1)

now  

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i)  by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c$.....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$.....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$