How do you use Integration by Substitution to find #intx^2*sqrt(x^3+1)dx#?

Question

Elijah Abram · Accepted Answer

Using integration by substitution, we find the answer to this problem to be #2/9(x^3+1)^(3/2)#.

We need to find a value for #u# to substitute that will allow us to find this integral more easily. Notice that if we choose our #u# to be #x^3+1#, then our #du# (the derivative of our #u#) will be #3x^2dx#. The degree of this answer matches up with what is outside of the radical—they only differ by a constant, which works for us since we can multiply by a constant as well as we divide by that same constant as well. So with these changes, the integral becomes #1/3int3x^2sqrt(x^3+1) dx#. Now we can substitute in #u# for #sqrt(x^3+1)# and #du# for #3x^2dx# (even though they're on different sides of the integral we can combine them into the #du#) and we are left with this, in terms of #u#: #1/3intsqrt(u) du# Now this is a problem that we can use the anti-power rule on: this integral becomes #1/3[2/3u^(3/2)]=2/9u^(3/2)#. A very important step that is often overlooked at the end of the problem is that because this is an indefinite integral, we need to substitute back in so that the final answer is in terms of the variable we started with (in this case, #x#). Remember that we substituted #u=x^3+1#, so our final answer becomes #2/9(x^3+1)^(3/2)#.

William Abdalla · Answer

undefined To solve the integral $ \int x^2 \sqrt{x^3 + 1} \, dx $ using integration by substitution:

1. Let $ u = x^3 + 1 $, then $ du = 3x^2 \, dx $.
2. Rewrite the integral in terms of $ u $: $ \int x^2 \sqrt{x^3 + 1} \, dx = \frac{1}{3} \int \sqrt{u} \, du $.
3. Integrate $ \sqrt{u} $ with respect to $ u $.
4. Substitute back $ x^3 + 1 $ for $ u $ in the final result.

So, after performing the substitution and integrating, you'll have the solution in terms of $ x $.