Solve $\int X^5 \sqrt{1+x^6} D X$ Using Substitution

Calculating indefinite integrals is a fundamental skill in calculus, and today, we're going to tackle a fascinating one: evaluate the indefinite integral of $\int x^5 \sqrt{1+x^6} d x$. This problem is a perfect candidate for the u-substitution method, a powerful technique that simplifies complex integrals by transforming them into simpler forms. We'll break down each step, making sure you understand the 'why' behind each move. So, grab your favorite note-taking tool, and let's dive into the world of calculus!

Understanding the U-Substitution Method

The u-substitution method is essentially the chain rule in reverse. When you encounter an integral that looks like it's made up of a function and its derivative (or a constant multiple of its derivative), u-substitution is often your best bet. The core idea is to replace a complex part of the integrand with a new variable, usually denoted by 'u'. This substitution simplifies the integral, making it easier to integrate. After integrating with respect to 'u', we substitute back the original expression in terms of 'x' to get our final answer. It's like giving a complex puzzle a simpler temporary name, solving the puzzle with the simpler name, and then remembering what the original complex pieces were. This method is incredibly versatile and forms the backbone of solving many integration problems. The key to successfully applying u-substitution lies in identifying the correct part of the integrand to set as 'u'. Often, this is the 'inner function' whose derivative also appears in the integrand. For our specific problem, $\int x^5 \sqrt{1+x^6} d x$, we'll need to carefully choose our 'u' to maximize the simplification.

Step-by-Step Solution: Evaluating $\int x^5 \sqrt{1+x^6} d x$

Let's begin the exciting journey of solving the indefinite integral $\int x^5 \sqrt{1+x^6} d x$. Our primary tool will be the u-substitution method. First, we need to make a strategic choice for our substitution. Observe the expression under the square root: $1+x^6$. Notice that its derivative is $6x^5$. And conveniently, we have an $x^5$ term multiplying the square root in our integrand. This strongly suggests setting $u = 1+x^6$. Let's proceed with this choice.

Step 1: Choose the substitution. Let $u = 1+x^6$.

Step 2: Find the differential $du$. To find $du$, we differentiate our expression for $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(1+x^6) = 6x^5$. Now, we rearrange this to solve for $du$: $du = 6x^5 dx$.

Step 3: Substitute into the integral. Our original integral is $\int x^5 \sqrt{1+x^6} d x$. We have $u = 1+x^6$, so $\sqrt{1+x^6}$ becomes $\sqrt{u}$. We also have $du = 6x^5 dx$. Notice that we have $x^5 dx$ in our integral. We can rewrite $du = 6x^5 dx$ as $\frac{1}{6} du = x^5 dx$. Now we can substitute these into the integral:

$\int \sqrt{1+x^6} (x^5 dx) = \int \sqrt{u} \left(\frac{1}{6} du\right)$

Step 4: Simplify and integrate with respect to $u$. We can pull the constant $\frac{1}{6}$ out of the integral:

$\frac{1}{6} \int \sqrt{u} du$

To integrate $\sqrt{u}$, we rewrite it in exponent form: $\sqrt{u} = u^{1/2}$. Now, we can use the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$). Applying this rule to $u^{1/2}$:

$\frac{1}{6} \int u^{1/2} du = \frac{1}{6} \left(\frac{u^{1/2 + 1}}{1/2 + 1}\right) + C$

$= \frac{1}{6} \left(\frac{u^{3/2}}{3/2}\right) + C$

$= \frac{1}{6} \left(\frac{2}{3} u^{3/2}\right) + C$

Now, we multiply the constants:

$= \frac{2}{18} u^{3/2} + C$

$= \frac{1}{9} u^{3/2} + C$

Step 5: Substitute back in terms of $x$. Our integral is now solved in terms of $u$, but the original problem was in terms of $x$. We must substitute back $u = 1+x^6$ into our result:

$= \frac{1}{9} (1+x^6)^{3/2} + C$

And there you have it! The solution to the indefinite integral $\int x^5 \sqrt{1+x^6} d x$ is $\frac{1}{9} (1+x^6)^{3/2} + C$. Remember that the '+ C' represents the constant of integration, which is essential for all indefinite integrals.

Verifying the Solution

To ensure our calculation is correct, we can verify the solution by differentiating our result with respect to $x$. If we get the original integrand, our integration was successful. Our integrated function is $F(x) = \frac{1}{9} (1+x^6)^{3/2} + C$. Let's find its derivative, $F'(x)$:

$F'(x) = \frac{d}{dx} \left(\frac{1}{9} (1+x^6)^{3/2} + C\right)$

We can pull the constant $\frac{1}{9}$ out and differentiate the $(1+x^6)^{3/2}$ term using the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. In our case, the outer function is $f(v) = v^{3/2}$ and the inner function is $g(x) = 1+x^6$.

First, find the derivative of the outer function with respect to $v$: $f'(v) = \frac{3}{2} v^{(3/2)-1} = \frac{3}{2} v^{1/2}$.

Next, find the derivative of the inner function with respect to $x$: $g'(x) = \frac{d}{dx}(1+x^6) = 6x^5$.

Now, apply the chain rule: $F'(x) = \frac{1}{9} \cdot \left(\frac{3}{2} (1+x^6)^{1/2}\right) \cdot (6x^5)$

Multiply the constants: $\frac{1}{9} \cdot \frac{3}{2} \cdot 6 = \frac{18}{18} = 1$.

So, the derivative becomes:

$F'(x) = 1 \cdot (1+x^6)^{1/2} \cdot x^5$

$F'(x) = x^5 \sqrt{1+x^6}$

This matches our original integrand $\int x^5 \sqrt{1+x^6} d x$. This confirms that our solution $\frac{1}{9} (1+x^6)^{3/2} + C$ is indeed correct. The process of differentiation acted as a perfect check for our integration steps.

Why U-Substitution Works: A Deeper Look

The u-substitution method is a cornerstone of integration, and its effectiveness stems directly from the chain rule of differentiation. Remember the chain rule: if you have a composite function $f(g(x))$, its derivative is $f'(g(x)) imes g'(x)$. When we see an integral that resembles this form, like $\int f(g(x)) g'(x) dx$, we can simplify it. By setting $u = g(x)$, we find that $du = g'(x) dx$. Substituting these into the integral gives us $\int f(u) du$. This transformed integral is often much easier to solve. After solving for $u$, we replace $u$ with $g(x)$ to get the answer in terms of the original variable. This technique is so powerful because it allows us to