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

Q: What is \int \frac{1}{x^2}\,dx ?

Rewrite \frac{1}{x^2}=x^{-2}. Then \int x^{-2}\,dx=\frac{x^{-1}}{-1}=-\frac{1}{x}+C.

Q: How do you integrate x^{-2}?

Use the power rule: \int x^n\,dx=\frac{x^{n+1}}{n+1}+C for n\neq -1. For n=-2: \int x^{-2}\,dx=-x^{-1}+C=-\frac{1}{x}+C.

Q: What is the derivative of -\frac{1}{x}?

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

Q: Is \int \frac{1}{x^2}\,dx valid for x=0?

No. The integrand \frac{1}{x^2} is undefined at x=0. So the antiderivative works on intervals not crossing 0.

Q: What is \int \frac{1}{x^2}\,dx using integration by parts?

Let u=\frac{1}{x^2}, dv=dx leads to a cycle. The easiest method is the power rule: \int x^{-2}\,dx=-\frac{1}{x}+C.

Q: Compute a definite integral like \int_{1}^{2}\frac{1}{x^2}\,dx .

Use -\frac{1}{x} : \left[-\frac{1}{x}\right]_{1}^{2}= -\frac{1}{2}-(-1)=\frac{1}{2}.

Answer

We want to compute

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

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

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

So the result is

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

Detailed Explanation

We want to find the indefinite integral

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

Step 1: Rewrite the integrand using exponents.

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

Step 2: Use the power rule for integration.

The power rule says that for \(n \neq -1\),

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

Here, \(n = -2\). Since \(-2 \neq -1\), the rule applies.

Step 3: Apply the power rule.

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

Step 4: Simplify the exponents and the denominator.

\[ -2+1=-1, \quad \text{and} \quad \frac{1}{-1}=-1. \]

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

Step 5: Rewrite \(x^{-1}\) as a fraction.

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

So,

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

Final Answer:

\[ -\frac{1}{x} + C. \]

See full solution

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Calculus FAQ

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

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

How do you integrate \(x^{-2}\)?

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

What is the derivative of \(-\frac{1}{x}\)?

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

Is \( \int \frac{1}{x^2}\,dx \) valid for \(x=0\)?

No. The integrand \( \frac{1}{x^2}\) is undefined at \(x=0\). So the antiderivative works on intervals not crossing \(0\).

What is \( \int \frac{1}{x^2}\,dx \) using integration by parts?

Let \(u=\frac{1}{x^2}\), \(dv=dx\) leads to a cycle. The easiest method is the power rule: \( \int x^{-2}\,dx=-\frac{1}{x}+C\).

Compute a definite integral like \( \int_{1}^{2}\frac{1}{x^2}\,dx \).

Use \( -\frac{1}{x}\) : \( \left[-\frac{1}{x}\right]_{1}^{2}= -\frac{1}{2}-(-1)=\frac{1}{2}\).

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