Let's break this down step by step. The equation represents the solution of the integral $\int x \tan^{-1}(x) \, dx$ using integration by parts and some algebraic manipulation.

Step 1: The formula for integration by parts

The method of integration by parts is given by: $\int u \, dv = uv - \int v \, du$ Here, we choose:

• $u = \tan^{-1}(x)$ (because its derivative simplifies nicely),
• $dv = x \, dx$.

Step 2: Compute $u$ and $dv$ derivatives and integrals

1. If $u = \tan^{-1}(x)$, then: $du = \frac{1}{1 + x^2} \, dx.$
2. If $dv = x \, dx$, then: $v = \frac{x^2}{2}.$

Now substitute these into the integration by parts formula: $\int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} \, dx.$

Step 3: Simplify the remaining integral

The remaining integral is: $\int \frac{x^2}{2(1 + x^2)} \, dx.$ Factorize the numerator $x^2$ as $(x^2 + 1) - 1$, so: $x^2 = (x^2 + 1) - 1.$ This allows us to rewrite the integral as: $\int \frac{x^2}{2(1 + x^2)} \, dx = \int \frac{(x^2 + 1) - 1}{2(1 + x^2)} \, dx.$

Split this into two simpler terms: $\int \frac{(x^2 + 1)}{2(1 + x^2)} \, dx - \int \frac{1}{2(1 + x^2)} \, dx.$

Step 4: Solve each part

1. The first term simplifies because $\frac{x^2 + 1}{1 + x^2} = 1$: $\int \frac{(x^2 + 1)}{2(1 + x^2)} \, dx = \int \frac{1}{2} \, dx = \frac{x}{2}.$

2. The second term is: $\int \frac{1}{2(1 + x^2)} \, dx = \frac{1}{2} \int \frac{1}{1 + x^2} \, dx = \frac{1}{2} \tan^{-1}(x).$

Step 5: Combine everything

Combine the results into the original equation: $\int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \left( \frac{x}{2} - \frac{1}{2} \tan^{-1}(x) \right).$

Simplify: $\int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x).$

This process carefully handles the integral, simplifying the algebra and breaking the solution into manageable parts.

Questions:

1. How is integration by parts applied in general problems?
2. Why is $u = \tan^{-1}(x)$ a good choice for this problem?
3. How is the numerator $x^2$ rewritten as $(x^2 + 1) - 1$, and why is this useful?
4. What are the key simplifications made to compute the integral of $\frac{1}{1 + x^2}$?
5. Can this approach work for other similar integrals involving inverse trigonometric functions?

Tip:

When choosing $u$ and $dv$ in integration by parts, always pick $u$ as the term that simplifies upon differentiation.