Question

evaluate integral of.(3x²+2x-1)⁷(6x+2) dx​

Answer 1

Answer:

$\frac{ ({3 {x}^{2} + 2x - 1 })^{8} }{8} + c$

Step-by-step explanation:

Assume

$t = 3 {x}^{2} + 2x - 1$

So after differentiating, we get

$dt = (6x + 2)dx$

So, now we need the integral of

${t}^{7} dt$

Which equals,

$\frac{ {t}^{8} }{8}$

So, we get after resubstituting t,

$\frac{ ({3 {x}^{2} + 2x - 1 })^{8} }{8} + c$

where 'c' is the constant of integration.