Respuesta :
Answer:
[tex]log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x[/tex]
Step-by-step explanation:
step 1:- by using partial fractions
[tex][tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }[/tex]......(1)
step 2:-
solving on both sides
[tex]1-x=A(x^{2} +1)+(Bx+C)x......(2)[/tex]
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
[tex]\\\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[/tex]
by using integration formulas
i) by using [tex]\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\\[/tex].....(b)
[tex]\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C[/tex].....(c)
step 4:-
by using above integration formulas (a,b,and c)
we get answer is
[tex]log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x[/tex]
