## Problem Statement

Calculate the indefinite integral:

$$
\int \operatorname{arctg} \sqrt{3 x-1} \, d x
$$
## Solution

$$
\int \operatorname{arctg} \sqrt{3 x-1} d x=
$$

Let:

$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{3 x-1} ; d u=\frac{1}{1+(\sqrt{3 x-1})^{2}} \cdot \frac{1}{2 \sqrt{3 x-1}} \cdot 3 d x= \\
& =\frac{1}{3 x} \cdot \frac{3}{2 \sqrt{3 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{3 x-1}} \\
& d v=d x ; v=x
\end{aligned}
$$

Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:

$$
\begin{aligned}
& =\operatorname{arctg} \sqrt{3 x-1} \cdot x-\int_{1} \frac{x \cdot d x}{2 x \sqrt{3 x-1}}=x \cdot \operatorname{arctg} \sqrt{3 x-1}-\int \frac{d x}{2 \sqrt{3 x-1}}= \\
& =x \cdot \operatorname{arctg} \sqrt{3 x-1}-\frac{1}{3} \cdot \sqrt{3 x-1}+C
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_1-15" Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

- Last edited on this page: 21:15, 18 February 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 1-16

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