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If intsqrt((x-5)/(x-7))dx=Asqrt(x^(2)-12x+35) +log|x-6+sqrt(x^(2)-12x+35)|+C, then A= |
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Answer» `-1` `=intsqrt(((x-5)(x-5))/((x-7)(x-5)))dx` [rationalising] `=INT(x-5)/(sqrt(x^(2)-5x-7x+35))dx` `=int(x-5)/(sqrt(x^(2)-12x+25))dx` `=(1)/(2)int(2x-10)/(sqrt(x^(2)-12x+35))` `=(1)/(2)int(2x-12+2)/(sqrt(x^(2)_12x+35))dx` `=(1)/(2)int(2x-12)/(sqrt(x^(2)-12x+35))dx+int(1)/(sqrt(x^(2)-12x+35))dx` `=sqrt(x^(2)-12x+35)+int(1)/(sqrt(x^(2)-12x+36-1))dx+C` `[because" let "x^(2)-12x+35=timplies(2x-12)dx=dt]` `sqrt(x^(2)-12x+35)+int(1)/(sqrt((x-6)^(2)-1^(2)))dx+c` `l=sqrt(x^(2)-12x+35)log+|x-6+sqrt(x^(2)-12x+35)|+C` `thereforel=Asqrt(x^(2)-12x+35)+log|x-6+sqrt(x^(2)-12+35)+C` `=1*sqrt(x^(2)-12x+35)log|x-6+sqrt(x^(2)-12x+35)+C` On COMPARING both sides, we get A=1 |
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