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 for a constant multiple \(k\) of a function, \(\int kf(x)dx=k\int f(x)dx\). Here, the integrand is \(10x^{\sqrt{5}}\), where \(10\) is a constant and the power of \(x\) is \(n = \sqrt{5}\).

Step2: Apply the constant multiple rule and power rule

First, use the constant multiple rule: \(\int10x^{\sqrt{5}}dx=10\int x^{\sqrt{5}}dx\).
Then, apply the power rule \(\int x^n dx=\frac{x^{n + 1}}{n + 1}+C\) with \(n=\sqrt{5}\). So we have:
\(10\times\frac{x^{\sqrt{5}+1}}{\sqrt{5}+1}+C=\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C\)
We can also rationalize the denominator (though it's not necessary for the antiderivative, but for simplification):
Multiply numerator and denominator by \(\sqrt{5}-1\):
\(\frac{10(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}x^{\sqrt{5}+1}+C=\frac{10(\sqrt{5}-1)}{5 - 1}x^{\sqrt{5}+1}+C=\frac{10(\sqrt{5}-1)}{4}x^{\sqrt{5}+1}+C=\frac{5(\sqrt{5}-1)}{2}x^{\sqrt{5}+1}+C\)
But the simplest form using the power rule directly is \(\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C\) or after simplifying the coefficient, we can leave it as \(10\times\frac{x^{\sqrt{5}+1}}{\sqrt{5}+1}+C\)

Answer:

\(\frac{10}{\sqrt{5}+1}x^{\sqrt{5}+1}+C\) (or \(\frac{5(\sqrt{5}-1)}{2}x^{\sqrt{5}+1}+C\) after rationalizing the denominator)