Answer:
[tex]\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C[/tex]
tep-by-step explanation:
In order to find the integral:
[tex]\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx[/tex]
we can do the following substitution:
Let's call
[tex]u=(x^3-x^2+x)[/tex]
Then
[tex]du = (3x^2-2x+1) dx[/tex]
which allows us to do convert the original integral into a much simpler one of easy solution:
[tex]\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \int\ {u^7 \, du = \frac{1}{8} \,u^8 +C[/tex]
Therefore, our integral written in terms of "x" would be:
[tex]\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C[/tex]
