## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{4} \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{1}^{4} \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x=\int_{1}^{4} \frac{d(\sqrt{x}+x)}{(\sqrt{x}+x)^{2}}=-\left.\frac{1}{\sqrt{x}+x}\right|_{1} ^{4}= \\
& =-\frac{1}{\sqrt{4}+4}+\frac{1}{\sqrt{1}+1}=-\frac{1}{6}+\frac{1}{2}=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-8$ »

Categories: Kuznetsov Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals $4-9$

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