Math Snap

STEP 1

1. We are tasked with evaluating the integral $\int \frac{\operatorname{arctg} x}{x^{2}+1} \, dx$.
2. $\operatorname{arctg} x$ is the inverse tangent function, also denoted as $\tan^{-1} x$.

STEP 2

1. Consider a substitution to simplify the integral.
2. Apply integration by parts.
3. Simplify and solve the resulting integral.

STEP 3

Consider the substitution $u = \operatorname{arctg} x$.
Then, $du = \frac{1}{x^2 + 1} \, dx$.
This substitution simplifies the integral to:
$$\int u \, du$$

STEP 4

Apply integration by parts to the integral $\int u \, du$.
Recall the formula for integration by parts: $\int u \, dv = uv - \int v \, du$.
For our integral, choose:
- $u = \operatorname{arctg} x$, hence $du = \frac{1}{x^2 + 1} \, dx$
- $dv = \frac{1}{x^2 + 1} \, dx$, hence $v = \tan^{-1} x$

STEP 5

Calculate $v$ by integrating $dv$:
Since $dv = \frac{1}{x^2 + 1} \, dx$, we have:
$$v = \int \frac{1}{x^2 + 1} \, dx = \operatorname{arctg} x$$

STEP 6

Substitute into the integration by parts formula:
$$\int \frac{\operatorname{arctg} x}{x^2 + 1} \, dx = \operatorname{arctg} x \cdot \operatorname{arctg} x - \int \operatorname{arctg} x \cdot \frac{1}{x^2 + 1} \, dx$$ Notice that the second integral is the same as the original integral.

Solve the equation:
Let $I = \int \frac{\operatorname{arctg} x}{x^2 + 1} \, dx$.
Then:
$$I = (\operatorname{arctg} x)^2 - I$$ $$2I = (\operatorname{arctg} x)^2$$ $$I = \frac{1}{2} (\operatorname{arctg} x)^2 + C$$ where $C$ is the constant of integration.
The solution to the integral is:
$$\boxed{\frac{1}{2} (\operatorname{arctg} x)^2 + C}$$