Question

find the most general antiderivative or indefinite integral. \\(\int 10x^{\sqrt{5}} dx\\)

Explanation:

Step1: Recall the power rule for integration

The power rule for integration states that $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$, where $n
eq - 1$ and $C$ is the constant of integration. Also, we can factor out the constant coefficient from the integral, i.e., $\int k\cdot f(x)dx=k\int f(x)dx$ where $k$ is a constant.

For the integral $\int10x^{\sqrt{5}}dx$, we first factor out the constant $10$:
$\int10x^{\sqrt{5}}dx = 10\int x^{\sqrt{5}}dx$

Step2: Apply the power rule

Here, $n = \sqrt{5}$. Using the power rule $\int x^n dx=\frac{x^{n + 1}}{n + 1}+C$, we have:

$10\int x^{\sqrt{5}}dx=10\times\frac{x^{\sqrt{5}+1}}{\sqrt{5}+1}+C$

Simplify the expression:
$=\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C$

We can also rationalize the denominator of $\frac{10}{\sqrt{5}+1}$ by multiplying the numerator and denominator by $\sqrt{5}-1$:

$\frac{10(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}x^{\sqrt{5}+1}+C$

Since $(\sqrt{5}+1)(\sqrt{5}-1)=(\sqrt{5})^2-1^2 = 5 - 1=4$, the expression becomes:

$\frac{10(\sqrt{5}-1)}{4}x^{\sqrt{5}+1}+C=\frac{5(\sqrt{5}-1)}{2}x^{\sqrt{5}+1}+C$

But the form $\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C$ (or the simplified rationalized form) is also correct. The most straightforward form after applying the power rule without rationalizing is $\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C$ or we can write it as $\frac{10x^{\sqrt{5}+1}}{\sqrt{5}+1}+C$

Answer:

$\frac{10x^{\sqrt{5}+1}}{\sqrt{5}+1}+C$ (or the rationalized form $\frac{5(\sqrt{5}-1)}{2}x^{\sqrt{5}+1}+C$)