# Math Help - Integral. Verification of decision

1. ## Integral. Verification of decision

Can I find this integral by more simple method?
I think that I used bad replacement of the variable of integration.

$
\int {\frac{{dx}}
{{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} = \left\{ \begin{gathered}
x = t^6 , \hfill \\
dx = 6t^5 dt \hfill \\
\end{gathered} \right\} = 6\int {\frac{{dt}}
{{t\left( {2t^3 + t^2 + 1} \right)}}.}
$

$
\begin{gathered}
2t^3 + t^2 + 1 = 2t^3 + 2 + t^2 - 1 = 2\left( {t^3 + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \hfill \\
= 2\left( {t + 1} \right)\left( {t^2 - t + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \left( {t + 1} \right)\left( {2t^2 - t + 1} \right). \hfill \\
\end{gathered}
$

$\begin{gathered}
\frac{1}
{{t\left( {t + 1} \right)\left( {2t^2 - t + 1} \right)}} = \frac{A}
{t} + \frac{B}
{{t + 1}} + \frac{{Ct + D}}
{{2t^2 - t + 1}} \Leftrightarrow \hfill \\
\Leftrightarrow 1 = A\left( {t + 1} \right)\left( {2t^2 - t + 1} \right) + B\left( {2t^2 - t + 1} \right)t + \left( {Ct + D} \right)\left( {t + 1} \right)t = \hfill \\
\end{gathered}
$

$
\begin{gathered}
= 2At^3 + At^2 + A + 2Bt^3 - Bt^2 + Bt + Ct^3 + Ct^2 + Dt^2 + Dt = \hfill \\
= \left( {2A + 2B + C} \right)t^3 + \left( {A - B + C + D} \right)t^2 + \left( {B + D} \right)t + A{\text{ }} \Rightarrow \hfill \\
\end{gathered}
$

$
\Rightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B + D = 0, \hfill \\
2A + 2B + C = 0, \hfill \\
A - B + C + D = 0; \hfill \\
\end{gathered} \right.
$

$
\Leftrightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B = - D, \hfill \\
2D - C = 2, \hfill \\
2D + C = - 1; \hfill \\
\end{gathered} \right.
$

$
\Leftrightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B = - {1 \mathord{\left/
{\vphantom {1 {4,}}} \right.
\kern-\nulldelimiterspace} {4,}} \hfill \\
D = {1 \mathord{\left/
{\vphantom {1 4}} \right.
\kern-\nulldelimiterspace} 4}, \hfill \\
C = - {3 \mathord{\left/
{\vphantom {3 {2.}}} \right.
\kern-\nulldelimiterspace} {2.}} \hfill \\
\end{gathered} \right.
$

$
\begin{gathered}
6\int {\frac{{dt}}
{{t\left( {2t^3 + t^2 + 1} \right)}}} = 6\int {\frac{1}
{t}dt} - \frac{3}
{2}\int {\frac{1}
{{t + 1}}} dt - \frac{3}
{2}\int {\frac{{6t - 1}}
{{2t^2 - t + 1}}} dt = I_1 - I_2 - I_3 . \hfill \\
I_1 = 6\int {\frac{1}
{t}dt} = 6\ln t + C = 6\ln \left( {\sqrt[6]{x}} \right) + C = \ln x + C. \hfill \\
\end{gathered}
$

$
I_2 = \frac{3}
{2}\int {\frac{1}
{{t + 1}}} dt = \frac{3}
{2}\ln \left( {t + 1} \right) + C = \frac{3}
{2}\ln \left( {\sqrt[6]{x} + 1} \right) + C.
$

$
I_3 = \frac{3}
{2}\int {\frac{{6t - 1}}
{{2t^2 - t + 1}}} dt = \frac{9}
{4}\int {\frac{{4t - {2 \mathord{\left/
{\vphantom {2 3}} \right.
\kern-\nulldelimiterspace} 3}}}
{{2t^2 - t + 1}}dt} = \frac{9}
{4}\int {\frac{{4t - 1 + 1 - {2 \mathord{\left/
{\vphantom {2 3}} \right.
\kern-\nulldelimiterspace} 3}}}
{{2t^2 - t + 1}}dt} =$

$
\begin{gathered}
= \frac{9}
{4}\int {\frac{{4t - 1}}
{{2t^2 - t + 1}}} dt + \frac{3}
{4}\int {\frac{{dt}}
{{2t^2 - t + 1}}} . \hfill \\
\frac{9}
{4}\int {\frac{{4t - 1}}
{{2t^2 - t + 1}}dt} = \frac{9}
{4}\int {\frac{{d\left( {2t^2 - t + 1} \right)}}
{{2t^2 - t + 1}}} = \frac{9}
{4}\ln \left( {2t^2 - t + 1} \right) + C. \hfill \\
\end{gathered}
$

$
\frac{3}
{4}\int {\frac{{dt}}
{{2t^2 - t + 1}}} = 6\int {\frac{{dt}}
{{16t^2 - 8t + 1 + 7}}} = 6\int {\frac{{dt}}
{{\left( {4t - 1} \right)^2 + 7}}} =
$

$
= \frac{6}
{7}\int {\frac{{dt}}
{{\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
{{2\sqrt 7 }}\int {\frac{{d\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)}}
{{\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right) + C.
$

$
\begin{gathered}
I_3 = \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right) + \frac{9}
{4}\ln \left( {2t^2 - t + 1} \right) + C = \hfill \\
= \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
{{\sqrt 7 }}} \right) + \frac{9}
{4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) + C. \hfill \\
\end{gathered}
$

$
\begin{gathered}
\boxed{\int {\frac{{dx}}
{{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3}
{2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9}
{4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) - \hfill \\
- \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
{{\sqrt 7 }}} \right) + C. \hfill \\
\end{gathered}
$

2. Originally Posted by DeMath
Can I find this integral by more simple method?
I think that I used bad replacement of the variable of integration.

$
\int {\frac{{dx}}
{{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} = \left\{ \begin{gathered}
x = t^6 , \hfill \\
dx = 6t^5 dt \hfill \\
\end{gathered} \right\} = 6\int {\frac{{dt}}
{{t\left( {2t^3 + t^2 + 1} \right)}}.}
$

$
\begin{gathered}
2t^3 + t^2 + 1 = 2t^3 + 2 + t^2 - 1 = 2\left( {t^3 + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \hfill \\
= 2\left( {t + 1} \right)\left( {t^2 - t + 1} \right) + \left( {t - 1} \right)\left( {t + 1} \right) = \left( {t + 1} \right)\left( {2t^2 - t + 1} \right). \hfill \\
\end{gathered}
$

$\begin{gathered}
\frac{1}
{{t\left( {t + 1} \right)\left( {2t^2 - t + 1} \right)}} = \frac{A}
{t} + \frac{B}
{{t + 1}} + \frac{{Ct + D}}
{{2t^2 - t + 1}} \Leftrightarrow \hfill \\
\Leftrightarrow 1 = A\left( {t + 1} \right)\left( {2t^2 - t + 1} \right) + B\left( {2t^2 - t + 1} \right)t + \left( {Ct + D} \right)\left( {t + 1} \right)t= \hfill \\
\end{gathered}
$

$
\begin{gathered}
= 2At^3 + At^2 + A + 2Bt^3 - Bt^2 + Bt + Ct^3 + Ct^2 + Dt^2 + Dt = \hfill \\
= \left( {2A + 2B + C} \right)t^3 + \left( {A - B + C + D} \right)t^2 + \left( {B + D} \right)t + A{\text{ }} \Rightarrow \hfill \\
\end{gathered}
$

$
\Rightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B + D = 0, \hfill \\
2A + 2B + C = 0, \hfill \\
A - B + C + D = 0; \hfill \\
\end{gathered} \right.
$

$
\Leftrightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B = - D, \hfill \\
2D - C = 2, \hfill \\
2D + C = - 1; \hfill \\
\end{gathered} \right.
$

$
\Leftrightarrow \left\{ \begin{gathered}
A = 1, \hfill \\
B = - {1 \mathord{\left/
{\vphantom {1 {4,}}} \right.
\kern-\nulldelimiterspace} {4,}} \hfill \\
D = {1 \mathord{\left/
{\vphantom {1 4}} \right.
\kern-\nulldelimiterspace} 4}, \hfill \\
C = - {3 \mathord{\left/
{\vphantom {3 {2.}}} \right.
\kern-\nulldelimiterspace} {2.}} \hfill \\
\end{gathered} \right.
$

$
\begin{gathered}
6\int {\frac{{dt}}
{{t\left( {2t^3 + t^2 + 1} \right)}}} = 6\int {\frac{1}
{t}dt} - \frac{3}
{2}\int {\frac{1}
{{t + 1}}} dt - \frac{3}
{2}\int {\frac{{6t - 1}}
{{2t^2 - t + 1}}} dt = I_1 - I_2 - I_3 . \hfill \\
I_1 = 6\int {\frac{1}
{t}dt} = 6\ln t + C = 6\ln \left( {\sqrt[6]{x}} \right) + C = \ln x + C. \hfill \\
\end{gathered}
$

$
I_2 = \frac{3}
{2}\int {\frac{1}
{{t + 1}}} dt = \frac{3}
{2}\ln \left( {t + 1} \right) + C = \frac{3}
{2}\ln \left( {\sqrt[6]{x} + 1} \right) + C.
$

$
I_3 = \frac{3}
{2}\int {\frac{{6t - 1}}
{{2t^2 - t + 1}}} dt = \frac{9}
{4}\int {\frac{{4t - {2 \mathord{\left/
{\vphantom {2 3}} \right.
\kern-\nulldelimiterspace} 3}}}
{{2t^2 - t + 1}}dt} = \frac{9}
{4}\int {\frac{{4t - 1 + 1 - {2 \mathord{\left/
{\vphantom {2 3}} \right.
\kern-\nulldelimiterspace} 3}}}
{{2t^2 - t + 1}}dt} =$

$
\begin{gathered}
= \frac{9}
{4}\int {\frac{{4t - 1}}
{{2t^2 - t + 1}}} dt + \frac{3}
{4}\int {\frac{{dt}}
{{2t^2 - t + 1}}} . \hfill \\
\frac{9}
{4}\int {\frac{{4t - 1}}
{{2t^2 - t + 1}}dt} = \frac{9}
{4}\int {\frac{{d\left( {2t^2 - t + 1} \right)}}
{{2t^2 - t + 1}}} = \frac{9}
{4}\ln \left( {2t^2 - t + 1} \right) + C. \hfill \\
\end{gathered}
$

$
\frac{3}
{4}\int {\frac{{dt}}
{{2t^2 - t + 1}}} = 6\int {\frac{{dt}}
{{16t^2 - 8t + 1 + 7}}} = 6\int {\frac{{dt}}
{{\left( {4t - 1} \right)^2 + 7}}} =
$

$
= \frac{6}
{7}\int {\frac{{dt}}
{{\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
{{2\sqrt 7 }}\int {\frac{{d\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)}}
{{\left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right)^2 + 1}}} = \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right) + C.
$

$
\begin{gathered}
I_3 = \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4t - 1}}
{{\sqrt 7 }}} \right) + \frac{9}
{4}\ln \left( {2t^2 - t + 1} \right) + C = \hfill \\
= \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
{{\sqrt 7 }}} \right) + \frac{9}
{4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) + C. \hfill \\
\end{gathered}
$

$
\begin{gathered}
\boxed{\int {\frac{{dx}}
{{x\left( {1 + 2\sqrt x + \sqrt[3]{x}} \right)}}} } = \ln x - \frac{3}
{2}\ln \left( {\sqrt[6]{x} + 1} \right) - \frac{9}
{4}\ln \left( {2\sqrt[3]{x} - \sqrt[6]{x} + 1} \right) - \hfill \\
- \frac{3}
{{2\sqrt 7 }}\arctan \left( {\frac{{4\sqrt[6]{x} - 1}}
{{\sqrt 7 }}} \right) + C. \hfill \\
\end{gathered}
$
there's nothing wrong with your solution. good work!