Question Statement

Evaluate the integral:

$\int {\tan }^{-1}xdx$

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Background and Explanation

This integral involves an inverse trigonometric function that doesn't have an obvious antiderivative. We use integration by parts, treating ${\tan }^{-1}x$ as the first function and $1$ as the second function, to transform the problem into a more manageable form.

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Solution

Let $I=\int {\tan }^{-1}xdx$

We apply integration by parts using the formula: $\int uvdx=u\int vdx-\int \left(\frac{du}{dx}\cdot \int vdx\right)dx$

Step 1: Choose the functions

• First function: $u={\tan }^{-1}x$
• Second function: $v=1$

This gives us:

• $\int vdx=\int 1dx=x$
• $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$

Step 2: Apply the integration by parts formula

$I={\tan }^{-1}x\cdot \int 1dx-\int \left(\int 1dx\right)\times \frac{d}{dx}\left({\tan }^{-1}x\right)dx$

$={\tan }^{-1}x\cdot x-\int x\cdot \frac{1}{1+x^2}dx$

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

Step 3: Evaluate the remaining integral

Notice that the numerator $x$ is related to the derivative of the denominator $1+x^2$. Specifically, $\frac{d}{dx}(1+x^2)=2x$, so we need a factor of $2$ in the numerator. We multiply and divide by $2$:

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

Step 4: Apply the logarithmic integration formula

Using the property that $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))$ (where $f(x)=1+x^2$ and $f^{\prime }(x)=2x$):

$=x{\tan }^{-1}x-\frac{1}{2}\ln \left(1+x^2\right)+C$

where $C$ is the constant of integration.

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Key Formulas or Methods Used

• Integration by parts: $\int uvdx=u\int vdx-\int \left(\frac{du}{dx}\cdot \int vdx\right)dx$
• Derivative of inverse tangent: $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$
• Logarithmic integration: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))$ (or $\ln |f(x)|$)

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Summary of Steps

1. Set up integration by parts with $u={\tan }^{-1}x$ (first function) and $v=1$ (second function)
2. Apply the formula to obtain: $x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx$
3. Manipulate the remaining integral by factoring out $\frac{1}{2}$ to get $\frac{2x}{1+x^2}$ in the numerator
4. Integrate using the logarithmic rule $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln (f(x))$ to get $\frac{1}{2}\ln (1+x^2)$
5. Combine all terms and add the constant of integration $C$