I=∫{(7x-x2)/[(2-x)(1+x2)]}dx. (No limits given for a definite integral)
Express in partial fractions:
Let (7x-x2)/[(2-x)(1+x2)]=A/(2-x)+(Bx+C)/(1+x2);
7x-x2=A+Ax2+(Bx+C)(2-x)=A+Ax2+2Bx-Bx2+2C-Cx.
Equating coefficients:
constant: A+2C=0, C=-A/2,
x: 2B-C=7, 2B+A/2=7 (1)
x2: A-B=-1, 2A-2B=-2 (2)
(1)+(2)=5A/2=5, A=2⇒B=3, C=-1.
I=2∫dx/(2-x)+3∫(x/(1+x2))dx-∫dx/(1+x2),
I=-2ln(2-x)+3/2ln(1+x2)-tan-1(x).
If the limits were [0,1], then I=3/2ln(2)-π/4.