`int (x^(4) dx)/((x-1)(x^(2)+1))` का मान ज्ञात कीजिए।

Answer

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Explanation

सर्वप्रथम `int (x^(4) dx)/((x-1)(x^(2)+1))= int x^(4)/(x^(3)-x^(2)+x-1)`
`=(x+1)+1/(x^(3)-x^(2)+x-1)`
`=(x+1)+1/((x-1)(x^(2)+1))`
अब `(1)/((x-1)(x^(2)+1))=A/(x-1)+(Bx+C)/(x^(2)+1)` [आंशिक भिन्न के रूप में]
`=(A(x^(2)+1)+(Bx+C)(x-1))/((x-1)(x^(2)+1))`
`1=Ax^(2)+A+Bx^(2)-Bx+Cx-C`
`0.x^(2)+0.x+1=(A+B)x^(2)+(C-B)x+(A-C)`
दोनों पक्षों की तुलना करने पर,
`A+B=0` अर्थात `A=-B`
`C-B=0` अर्थात `B=C`
और `A-C=1`
`:. -C-C=1" "[ :' A=-C]`
`-2C=1` या `C=-1/2`
`:. A=1/2` तथा `B=-1/2`
इसलिए `(1)/((x-1)(x^(2)+1))=(1)/(2(x-1))=(1)/(2(x-1))-x/(2(x^(2)+1))-1/(2(x^(2)+1))`
और `x^(4)/((x-1)(x^(2)+1))=(x+1)+1/(2(x-1))-x/(2(x^(2)+1))-1/(2(x^(2)+1))-1/(2(x^(2)+1))`
`:. int x^(4)/((x-1)(x^(2)+1)) dx= int (x+1) dx+1/2 int (dx)/(x-1) -1/2 int x/(x^(2)+1) dx - 1/2 int (dx)/(x^(2)+1)`
`=x^(2)/2+x+1/2 log|x-1|-1/4 log (x^(2)+1)-1/2 tan^(-1) x+c`.