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प्रश्न
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
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उत्तर
`( 3x + 5)sqrt(5 + 4x - 2x^2) dx`
Let 3x + 5 = A(4 - 4x) + B
⇒ A = `-3/4, B = 8`
I = `3/4(4 - 4x)sqrt(5 + 4x + 2x^2) dx + 8 sqrt(5 + 4x - 2x^2)dx`
= `-3/4 I_1 + 8l_2 ("let")`
For I1, put 5 + 4x = - 2x2 = t
⇒ (4 - 4x) dx = dt
`-3/4 I_1 = - 3/4 sqrtt dt = - 3/4 xx 2/3 t^(3/2)`
= `-1/2 (5 + 4x - 2x^2 )^(3/2)`
`8I_2 = 8sqrt2 sqrt(7/2 - ( x - 1)^2) dx`
I = `-1/2 (5 + 4x - 2x^2 )^(3/2) + 4sqrt2(x - 1) sqrt(5/2 + 2x - x^2) + 14sqrt2 sin^-1 (sqrt2(x - 1))/sqrt7 + C`
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