integrations plz solve

Answers
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


