## Solution

$$
\int_{-1 / 2}^{0} \frac{x d x}{2+\sqrt{2 x+1}}=
$$

Perform the substitution:

$$
t^{2}=2 x+1 ; d x=t d t
$$

Recalculate the limits of integration:

$$
\begin{aligned}
& x=-\frac{1}{2} \Rightarrow t=0 \\
& x=0 \Rightarrow t=1
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& \frac{1}{2} \int_{0}^{1} \frac{\left(t^{2}-1\right) t d t}{2+t}= \\
& =\frac{1}{2} \int_{0}^{1}\left(t^{2}-2 t+3-\frac{6}{t+2}\right) d t= \\
& =\left.\frac{1}{2}\left(\frac{t^{3}}{3}-2 \frac{t^{2}}{2}+3 t-6 \ln |t+2|\right)\right|_{0} ^{1}= \\
& =\frac{1}{2}\left(\frac{1}{3}-1+3-6 \ln 3+6 \ln 2\right)= \\
& =\frac{1}{2}\left(\frac{7}{3}-6 \ln \frac{3}{2}\right)=\frac{7}{6}-3 \ln \frac{3}{2}
\end{aligned}
$$

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_11-13” Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last edited on this page: 07:24, 28 June 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 11-14

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{-1 / 2}^{0} \frac{x \cdot d x}{2+\sqrt{2 x+1}}
$$