## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{64} \frac{6-\sqrt{x}+\sqrt[4]{x}}{\sqrt{x^{3}}-7 x-6 \sqrt[4]{x^{3}}} d x
$$
## Solution

$$
\int_{1}^{64} \frac{6-\sqrt{x}+\sqrt[4]{x}}{\sqrt{x^{3}}-7 x-6 \sqrt[4]{x^{3}}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=t^{4}, d x=4 t^{3} d t \\
& x=1 \Rightarrow t=\sqrt[4]{1}=1 \\
& x=64 \Rightarrow t=\sqrt[4]{64}=2 \sqrt{2}
\end{aligned}
$$

We get:

$$
=\int_{1}^{2 \sqrt{2}} \frac{\left(6-t^{2}+t\right) \cdot 4 t^{3} d t}{t^{6}-7 t^{4}-6 t^{3}}=-4 \cdot \int_{1}^{2 \sqrt{2}} \frac{t^{2}-t-6}{t^{3}-7 t-6} d t=
$$

Divide the denominator and numerator by \( t^{2}-t-6 \):

$$
\begin{aligned}
& \begin{array}{ll}
t^{3}-7 t-6 & \mid t^{2}-t-6 \\
t^{3}-t^{2}-6 t \\
\hline t^{2}-t-6 & t+1
\end{array} \\
& \frac{t^{2}-t-6}{0}
\end{aligned}
$$

We get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-28$ "

Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

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## Problem Kuznetsov Integrals 11-29

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