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Integrations plz solve |
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Answer»
To find ----> ∫ ( 3x - 2 ) √( 2x² - x - 1 ) dx Solution-----> 1) plzz see the attachment . 2) First we DIFFERENTIATE expression in bracket , d/dx ( 2x² - x + 1 ) = 4x - 1 We GET ( 4x - 1 ) , now , we FORM ( 4x - 1 ) in first bracket by taking 3 common first and then multiply and divide by 4 and then adding and subtracting 1 in first bracket. 3) now we get ( 4x - 1 ) in first bracket , now we break given intregation in to two intregations , 4) For first integration , we substitute 2x² - x + 1 = t Differentiating it we get, ( 4x - 1 ) dx = dt And then applying formula of intregation ∫ xⁿ dx = xⁿ⁺¹ / ( n + 1 ) + C 5) In second intregation we complete the whole square by adding and subtracting 1/16 . 6) Now we do a substitution x - ( 1 / 4 ) = u Differentiating both sides , we get, dx = du 7) Then we apply a formula of intregation. ∫ √x² + a² dx = x/2 √(x² + a² ) + a²/2 log ( x + √x² + a² ) + C
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