Question

Can please someone help me with this calculus problem? Thanks! ∫(4x-6)(x²-3x+1)^5 dx

Answer 1

Looks like the integral is

`I=\displaystyle\int (4x-6)(x^2-3x+1)^5\,\mathrm dx`

If you we substitute
`y=x^2-3x+1` and
`\mathrm dy=(2x-3)\,\mathrm dx`, we get

`I=\displaystyle2\int(2x-3)(x^2-3x+1)^5\,\mathrm dx=2\int y^5\,\mathrm dy`

Now use the power rule, and replace y accordingly:

`I=\frac{2y^6}6+C=\frac{y^6}3+C=\boxed{\frac{(x^2-3x+1)^5}3+C}`