Question

integrations plz solve​

Answer 1

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