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May I please get the full solution of this question number 16. It's urgent!
put x^2=y. so the expression now becomes[y+1]/[(y+2)(2y+1)]as the degree of the numerator is less than the denminator, we know that this can be broken into partial fractions as follows:[y+1]/[(y+2)(2y+1)]= A/(y+2)+B/(2y+1)to find the value of A and B, cross multiply and put y= – 2 and – ½ successively. so we obtain A=B=1/3so, replacing y with x^2, we have [x^2+1]/[(x^2+2)(2x^2+1)]= 1/3(x^2+2)+1/3(2x^2+1)so option B is correct
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