Q. \[ \int x^{-1/2}\,dx \]

Q: What substitution works for \int x^{-1/2} dx ?

Let x = u^2. Then dx = 2u\,du and x^{-1/2}=1/u. Integral becomes \int \frac{1}{u}\cdot 2u\,du = 2u + C = 2\sqrt{x}+C.

Q: How do you use the power rule when the exponent is negative?

Use \int x^n dx = \frac{x^{n+1}}{n+1} + C for n \ne -1. Here n=-\frac{1}{2}, so n+1=\frac{1}{2}. Thus \frac{x^{1/2}}{1/2}=2x^{1/2}=2\sqrt{x}.

Q: Why is there no special case like \int \frac{1}{x} dx ?

The special case is for exponent n=-1: \int x^{-1}dx=\ln|x|+C. Here n=-\frac{1}{2}\ne -1, so the power rule applies.

Q: What is the derivative check for the result 2\sqrt{x}+C ?

Differentiate: \frac{d}{dx}\left(2x^{1/2}\right)=2\cdot\frac{1}{2}x^{-1/2}=x^{-1/2}. So it matches the integrand.

Q: What domain restrictions apply for x^{-1/2} ?

For real values, require x>0 since x^{-1/2} = 1/\sqrt{x}. The antiderivative 2\sqrt{x}+C is valid on that domain.

Answer

We want to compute \(\int x^{-1/2}\,dx\).

Use the power rule \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) for \(n\neq -1\). Here \(n=-\dfrac{1}{2}\), so \(n+1=\dfrac{1}{2}\).

\[
\int x^{-1/2}\,dx=\frac{x^{1/2}}{1/2}+C=2x^{1/2}+C
\]

Final result: \(\,2\sqrt{x}+C\).

Detailed Explanation

We want to find the indefinite integral

\[
\int x^{-1/2}\,dx.
\]

Step 1: Identify the integration rule.

There is a power rule for integrals:

\[
\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C \quad \text{for } n\neq -1.
\]

Step 2: Match the exponent.

Here, the exponent is

\[
n=-\frac{1}{2}.
\]

Step 3: Verify the rule applies.

The rule requires \(n\neq -1\). Since \( -\frac{1}{2}\neq -1\), we can use the power rule.

Step 4: Add \(1\) to the exponent.

\[
n+1=-\frac{1}{2}+1=-\frac{1}{2}+\frac{2}{2}=\frac{1}{2}.
\]

Step 5: Divide by the new exponent.

Using the power rule:

\[
\int x^{-1/2}\,dx=\frac{x^{1/2}}{1/2}+C.
\]

Step 6: Simplify the coefficient.

Since \(\frac{1}{1/2}=2\), we get

\[
\int x^{-1/2}\,dx=2x^{1/2}+C.
\]

Step 7: Rewrite using a square root (optional but standard).

\[
x^{1/2}=\sqrt{x}.
\]

Final answer:

\[
\int x^{-1/2}\,dx=2\sqrt{x}+C.
\]

See full solution

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Calculus FAQ

What is \( \int x^{-1/2} \, dx \) ?

\(\int x^{-1/2} dx = \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + C\).

What substitution works for \( \int x^{-1/2} dx \) ?

Let \(x = u^2\). Then \(dx = 2u\,du\) and \(x^{-1/2}=1/u\). Integral becomes \(\int \frac{1}{u}\cdot 2u\,du = 2u + C = 2\sqrt{x}+C\).

How do you use the power rule when the exponent is negative?

Use \( \int x^n dx = \frac{x^{n+1}}{n+1} + C\) for \(n \ne -1\). Here \(n=-\frac{1}{2}\), so \(n+1=\frac{1}{2}\). Thus \(\frac{x^{1/2}}{1/2}=2x^{1/2}=2\sqrt{x}\).

Why is there no special case like \( \int \frac{1}{x} dx \) ?

The special case is for exponent \(n=-1\): \(\int x^{-1}dx=\ln|x|+C\). Here \(n=-\frac{1}{2}\ne -1\), so the power rule applies.

What is the derivative check for the result \(2\sqrt{x}+C\) ?

Differentiate: \( \frac{d}{dx}\left(2x^{1/2}\right)=2\cdot\frac{1}{2}x^{-1/2}=x^{-1/2}\). So it matches the integrand.

What domain restrictions apply for \(x^{-1/2}\) ?

For real values, require \(x>0\) since \(x^{-1/2} = 1/\sqrt{x}\). The antiderivative \(2\sqrt{x}+C\) is valid on that domain.

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