Q. \(\int x^{-3}\,dx\)

Q: How do you use the power rule for \int x^n\,dx when n=-3 ?

For n\neq -1 , \int x^n dx = \frac{x^{n+1}}{n+1}+C . With n=-3 , n+1=-2 , so \frac{x^{-2}}{-2}+C

Q: What is the derivative check for -\frac{1}{2x^{2}}+C ?

Differentiate: \frac{d}{dx}\left(-\frac{1}{2}x^{-2}\right)= -\frac{1}{2}\left(-2x^{-3}\right)=x^{-3}

Q: Is there a special case when the exponent is -1 , and does it apply here?

The special case is \int x^{-1}dx=\ln|x|+C . Here the exponent is -3 , so the power rule applies, not the log case

Q: What is the definite integral \int_{1}^{2} x^{-3}\,dx ?

Antiderivative: -\frac{1}{2x^{2}} . Evaluate: \left(-\frac{1}{8}\right)-\left(-\frac{1}{2}\right)=\frac{3}{8}

Q: What about integrating \int \frac{1}{x^{3}}\,dx ?

Since \frac{1}{x^{3}}=x^{-3} , \int \frac{1}{x^{3}}dx = -\frac{1}{2x^{2}}+C

Answer

To integrate \(x^{-3}\), use the power rule \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) for \(n\neq -1\).

\[
\int x^{-3}\,dx=\int x^{-3}\,dx=\frac{x^{-3+1}}{-3+1}+C=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^{2}}+C
\]

Final result: \(-\dfrac{1}{2x^{2}}+C\)

Detailed Explanation

We want to find the indefinite integral

\[
\int x^{-3}\,dx
\]

Step 1: Identify the rule to use

For any real number \(n \neq -1\), the power rule for integration is

\[
\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C
\]

Step 2: Match the problem to the rule

Here, \(n=-3\). This is allowed because \(-3 \neq -1\).

Step 3: Apply the power rule

Substitute \(n=-3\) into the formula:

\[
\int x^{-3}\,dx=\frac{x^{-3+1}}{-3+1}+C
\]

Step 4: Simplify the exponents and denominator

Compute the new exponent and denominator:

\[
-3+1=-2
\]

and

\[
-3+1=-2
\]

So the integral becomes

\[
\int x^{-3}\,dx=\frac{x^{-2}}{-2}+C
\]

Step 5: Rewrite in a more standard form

Since \(x^{-2}=\frac{1}{x^{2}}\), we can write

\[
\frac{x^{-2}}{-2}=-\frac{1}{2x^{2}}
\]

Final answer

\[
\int x^{-3}\,dx=-\frac{1}{2x^{2}}+C
\]

See full solution

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Calculus FAQ

What is the indefinite integral \( \int x^{-3}\,dx \)?

\( \int x^{-3}\,dx = \frac{x^{-2}}{-2}+C = -\frac{1}{2x^{2}}+C \)

How do you use the power rule for \( \int x^n\,dx \) when \( n=-3 \)?

For \( n\neq -1 \), \( \int x^n dx = \frac{x^{n+1}}{n+1}+C \). With \( n=-3 \), \( n+1=-2 \), so \( \frac{x^{-2}}{-2}+C \)

What is the derivative check for \( -\frac{1}{2x^{2}}+C \)?

Differentiate: \( \frac{d}{dx}\left(-\frac{1}{2}x^{-2}\right)= -\frac{1}{2}\left(-2x^{-3}\right)=x^{-3} \)

Is there a special case when the exponent is \( -1 \), and does it apply here?

The special case is \( \int x^{-1}dx=\ln|x|+C \). Here the exponent is \( -3 \), so the power rule applies, not the log case

What is the definite integral \( \int_{1}^{2} x^{-3}\,dx \)?

Antiderivative: \( -\frac{1}{2x^{2}} \). Evaluate: \( \left(-\frac{1}{8}\right)-\left(-\frac{1}{2}\right)=\frac{3}{8} \)

What about integrating \( \int \frac{1}{x^{3}}\,dx \)?

Since \( \frac{1}{x^{3}}=x^{-3} \), \( \int \frac{1}{x^{3}}dx = -\frac{1}{2x^{2}}+C \)

We solve the integral of x^-3.
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