## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{64} \frac{1-\sqrt[6]{x}+2 \sqrt[3]{x}}{x+2 \sqrt{x^{3}}+\sqrt[3]{x^{4}}} d x
$$
## Solution

$$
\int_{1}^{64} \frac{1-\sqrt[6]{x}+2 \sqrt[3]{x}}{x+2 \sqrt{x^{3}}+\sqrt[3]{x^{4}}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=t^{6}, d x=6 t^{5} d t \\
& x=1 \Rightarrow t=\sqrt[6]{1}=1 \\
& x=64 \Rightarrow t=\sqrt[6]{64}=2
\end{aligned}
$$

We get:

$$
=\int_{1}^{2} \frac{\left(1-t+2 t^{2}\right) \cdot 6 t^{5} d t}{t^{6}+2 t^{9}+t^{8}}=6 \cdot \int_{1}^{2} \frac{2 t^{2}-t+1}{2 t^{4}+t^{3}+t} d t=
$$

Divide the denominator and numerator by \(2 t^{2}-t+1\):

$$
\begin{aligned}
& \begin{array}{ll}
2 t^{4}+t^{3}+t & \mid 2 t^{2}-t+1 \\
\frac{2 t^{4}-t^{3}+t^{2}}{2 t^{3}-t^{2}}+t & t^{2}+t
\end{array} \\
& \frac{2 t^{3}-t^{2}+t}{0}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =6 \cdot \int_{1}^{2} \frac{1}{t^{2}+t} d t=6 \cdot \int_{1}^{2} \frac{1}{t(t+1)} d t=6 \cdot \int_{1}^{2}\left(\frac{1}{t}-\frac{1}{t+1}\right) d t= \\
& =\left.6 \cdot(\ln |t|-\ln |t+1|)\right|_{1} ^{2}=6 \cdot(\ln |2|-\ln |2+1|)-6 \cdot(\ln |1|-\ln |1+1|)= \\
& =6 \ln 2-6 \ln 3-6 \cdot 0+6 \ln 2=6 \ln \frac{4}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-2$ »

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

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