# 微分積分学‎: 積分: べき乗の置換積分

1.
$$\int \frac{2}{3+\sqrt[3]{x}} \, dx$$
解答
2.
$$\int x\sqrt[5]{x-1} \, dx$$
解答
3.
$$\int \frac{1}{2+\sqrt{x}} \, dx$$
解答
4.
$$\int \frac{5+\sqrt{x}}{2-\sqrt{x}} \, dx$$
解答
5.
$$\int \sqrt{3+\sqrt{x}} \, dx$$
解答
6.
$$\int x\sqrt[3]{2-x} \, dx$$
解答
7.
$$\int \sqrt[4]{1+\sqrt{x}} \, dx$$
解答
8.
$$\int \frac{1}{\sqrt[4]{x}+3} \, dx$$
解答

# Power Substitution - Introduction

To use Power Substitution, you should be familiar with the Integration by Substitution (or U-Substitution) and derivatives. The key thing to remember is:
If
$$u=f(x)$$
, then
$$du=f'(x) \, dx$$
.
For example:
If
$$u=5{x}^{3}-4x+5$$
, then
$$du={15x}^{2}-4 \, dx$$
.
The goal of power substitution is to replace any nth root (which is difficult to integrate), with integer power, which is much easier to integrate.

# Using Power Substitution

Let's illustrate this method using an example:
$$\int \frac{1}{2+\sqrt{x}} \, dx$$
Since this integral has an nth root, we decide to use Power Substitution. Let's substitute the term that has the nth root with
$$u$$
, then solve for
$$u$$
, and finally, find its derivative:
Let $$u=\sqrt{x}$$, $$x={u}^{2}$$ and $$dx=2u \, du$$
Now, let's substitute
$$u$$
into our original integral:
$$\int \frac{1}{2+u}\times 2u \, du$$
After simplifying, we have:
$$2\int \frac{u}{2+u} \, du$$
Now we have an easier integral without any square root. You can proceed to solve this. Just don't forget to substitute back
$$u=\sqrt{x}$$
after the integration.
We will not actually solve the integral in this tutorial. For the full answer with steps, see the full solution here. After all, the key is to recognize when to use power substitutions and what to set the variable
$$u$$
to. Hope you now have a good idea on how to do these.

# What's Next

At Cymath, we believe that learning by examples is one of the best ways to get better in calculus and problem solving in general. You can now try solving other integrals at the top of this page using Power Substitution. You can also subscribe to Cymath Plus, which offers ad-free and more in-depth help, from pre-algebra to calculus.