5 question answer please
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dx/x²-2x+4-(1)²+(1)²
=dx/(x²-2x+(1)²+(4-1)
=dx/(x-1)²+√3²
1/√3[tan^-1(x-1)/√3]
=1/√3[tan^-1{2-1/√3}-tan^-1{1-1/√3}]
1/√3[tan^-1{tanπ/6}-tan^-1(tan0)]
1/√3[π/6-0]
=π/6√3
=dx/(x²-2x+(1)²+(4-1)
=dx/(x-1)²+√3²
1/√3[tan^-1(x-1)/√3]
=1/√3[tan^-1{2-1/√3}-tan^-1{1-1/√3}]
1/√3[tan^-1{tanπ/6}-tan^-1(tan0)]
1/√3[π/6-0]
=π/6√3
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