How do you integrate #int x(x+1)^8# using substitution?

Question

Cameron Alexander · Accepted Answer

#(x+1)^10/10-(x+1)^9/9+C# Substitute #u=x+1# and you get: #int(u-1)u^8du# #=int u^9du - int u^8du# #=u^10/10-u^9/9+C# Substituting back #u=x+1# gives the answer.

Emily Ammerman · Answer

undefined To integrate ∫x(x+1)^8 using substitution, follow these steps:

1. Let u = x + 1.
2. Find du/dx. In this case, du/dx = 1.
3. Solve for dx in terms of du: dx = du.
4. Substitute u = x + 1 and dx = du into the integral.
5. Rewrite the integral in terms of u.
6. Integrate the new expression with respect to u.
7. Substitute back the expression involving x.

Let's do the steps:

1. Let $ u = x + 1 $.
2. Find $ \frac{du}{dx} $. In this case, $ \frac{du}{dx} = 1 $.
3. Solve for $ dx $ in terms of $ du $: $ dx = du $.
4. Substitute $ u = x + 1 $ and $ dx = du $ into the integral:
$$ \int x(x+1)^8 \, dx = \int (u-1)u^8 \, du $$
5. Rewrite the integral in terms of $ u $:
$$ \int (u-1)u^8 \, du = \int (u^9 - u^8) \, du $$
6. Integrate the new expression with respect to $ u $:
$$ \int (u^9 - u^8) \, du = \frac{u^{10}}{10} - \frac{u^9}{9} + C $$
7. Substitute back the expression involving $ x $:
$$ \frac{(x+1)^{10}}{10} - \frac{(x+1)^9}{9} + C $$