Question

Find the indefinite integral. (Use C for the constant of integration.)

Answer 1

Answer:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C$

tep-by-step explanation:

In order to find the integral:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx$

we can do the following substitution:

Let's call

$u=(x^3-x^2+x)$

Then

$du = (3x^2-2x+1) dx$

which allows us to do convert the original integral into a much simpler one of easy solution:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \int\ {u^7 \, du = \frac{1}{8} \,u^8 +C$

Therefore, our integral written in terms of "x" would be:

$\int\ {(3x^2-2x+1)\,(x^3-x^2+x)^7} \, dx = \frac{1}{8} (x^3-x^2+x)^8+C$