Question Statement

Evaluate the integral:

$\int \frac{{\cot }^{-1}x}{1+x^2}dx$

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Background and Explanation

This problem is solved using the method of integration by substitution (u-substitution). The key insight is recognizing that the derivative of ${\cot }^{-1}x$ is $\frac{-1}{1+x^2}$, which appears as a factor in the integrand.

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Solution

Let $I$ denote the integral we need to evaluate:

$I=\int \frac{{\cot }^{-1}x}{1+x^2}dx$

Step 1: Choose the substitution

Notice that the integrand contains ${\cot }^{-1}x$ and the term $\frac{1}{1+x^2}$. Since the derivative of ${\cot }^{-1}x$ involves $\frac{1}{1+x^2}$, this suggests the substitution:

$t={\cot }^{-1}x$

Step 2: Find the differential

Differentiating both sides with respect to $x$:

$\frac{dt}{dx}=\frac{-1}{1+x^2}$

Rearranging to express $dx$ in terms of $dt$:

$\frac{-1}{1+x^2}dx=dt$

Multiplying both sides by $-1$:

$\frac{1}{1+x^2}dx=-dt$

Step 3: Substitute into the integral

Replace ${\cot }^{-1}x$ with $t$ and $\frac{1}{1+x^2}dx$ with $-dt$:

$I=\int t\cdot (-dt)$

$I=-\int tdt$

Step 4: Integrate

Using the power rule for integration:

$I=-\frac{t^2}{2}+c$

Step 5: Back-substitute

Replace $t$ with ${\cot }^{-1}x$ to express the final answer in terms of $x$:

$I=\frac{-{\left({\cot }^{-1}x\right)}^2}{2}+c$

Or equivalently:

$I=-\frac{1}{2}{\left({\cot }^{-1}x\right)}^2+c$

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Key Formulas or Methods Used

• Integration by substitution: Used to simplify the integral by changing the variable
• Derivative of inverse cotangent: $\frac{d}{dx}({\cot }^{-1}x)=\frac{-1}{1+x^2}$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ (where $n\ne -1$)

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Summary of Steps

1. Set up substitution: Let $t={\cot }^{-1}x$
2. Differentiate: Find that $\frac{1}{1+x^2}dx=-dt$
3. Transform integral: Substitute to get $I=-\int tdt$
4. Integrate: Apply power rule to obtain $I=-\frac{t^2}{2}+c$
5. Substitute back: Replace $t$ to get final result $-\frac{({\cot }^{-1}x{)}^2}{2}+c$