Find `int_0^1 (7x-x^2)/((2-x)(x^2+1))dx`

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Let us write a partial fraction

`(7x-x^2)/((2-x)(x^2+1))=A/(2-x)+(Bx+c)/(x^2+1)`

`7x-x^2=A(x^2+1)+(Bx+c)(2-x)`

comparing the coefficients of `x^2,x and constant ` both side.

A-B=-1

2B-c=7

A+2c=0

solving the above system of equation we have

A=2,B=3 and c=-1

Thus

`int_1^1(7x-x^2)/((2-x)(x^2+1))dx=int_0^1{2/(2-x)+(3x-1)/(x^2+1)}dx`

`=int_0^1 2/(2-x)dx+int_0^1(3x)/(x^2+1)dx-int_0^1 1/(x^2+1)dx`

`` `=-ln(2-x)|^1_0+(3/2)int_0^1 (2x)/(x^2+1)dx-tan^(-1)(x)|^1_0`

`=-(ln(2-1)-ln(2-0))+(3/2){ln(x^2+1)|^1_0-(tan^(-1)(1)-tan^(-1)(0))`

`=-(0-ln(2))+(3/2)(ln(1+1)-ln(0+1))-pi/4`

`=ln(2)+(3/2)ln(2)-pi/4`

`=(1+3/2)ln(2)-pi/4`

`=(5/2)ln(2)-pi/4`

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