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प्रश्न
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
योग
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उत्तर
Let I = `int x^2/[ sqrt( x - 1) ] `dx
Substituting x - 1 = t and dx = dt, we get,
I = `int ( t + 1)^2/sqrt t`dx
= `int ( t^2 + 1 +2t )/ sqrtt` dt
= `int ( t^(3/2) + t^(-1/2) + 2t^(-1/2) )`dt
= `2/5t^(5/2) + 2t^(1/2) + 4/3t^(3/2)` + c
= `[ 6t^(5/2) + 30t^(1/2) + 20t^(3/2)]/15`+ c
= `2/15 t^(1/2)( 3t^2 + 15 + 10t )` + c
= `2/15 sqrt( x - 1 )[ 3( x -1 )^2 + 15 + 10( x - 1)]`+ c
= `2/15 sqrt( x - 1 )[ 3( x^2 + 1 - 2x ) + 15 + 10x - 10]`+ c
= `2/15 sqrt( x - 1 )[ 3x^2 + 3 - 6x + 15 + 10x - 10]`+ c
= `2/15 sqrt( x - 1 ) [ 3x^2 + 4x + 8 ]`+ c
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