Integrate the function in Exercise: x^ 2 +1 [ (x^ 2 +1)-2 ] x^ 4

Question

Integrate the function in Exercise:

$\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}$

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Answer

$\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\big[\log\text{(x}^{2}+1)-\log\text{x}^{2}\big]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\bigg[\log\bigg(\frac{\text{x}^{2}+1}{\text{x}^{2}}\bigg)\bigg]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{\text{x}^{2}+1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$\text{Let}\ 1+\frac{1}{\text{x}^{2}}=\text{t}\Rightarrow\frac{-2}{\text{x}^{3}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)\text{dx}$
$=-\frac{1}{2}\int\sqrt{\text{t}}\log\text{t dt}$
$=-\frac{1}{2}\int\text{t}^{\frac{1}{2}}.\log\text{t dt}$
Integrating by parts, we obtain
$\text{I}=-\frac{1}{2}\Bigg[\log\text{t}.\int\text{t}^{\frac{1}{2}}\text{dt}-\left\{\bigg(\frac{\text{d}}{\text{dt}}\log\text{t}\bigg)\int\text{t}^{\frac{1}{2}}\text{dt}\right\}\text{dt}\Bigg]$
$=-\frac{1}{2}\begin{bmatrix}\log\text{t}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}-\int\frac{1}{\text{t}}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}\text{dt} \\ \end{bmatrix}$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{2}{3}\int\text{t}^{\frac{1}{2}}\text{dt}\Bigg]$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{4}{9}\int\text{t}^{\frac{3}{2}}\text{dt}\Bigg]$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\log\text{t}+\frac{2}{9}\text{t}^{\frac{3}{2}}$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\bigg[\log\text{t}-\frac{2}{3}\bigg]$
$=-\frac{1}{3}\bigg(1+\frac{1}{\text{x}^{2}}\bigg)^{\frac{3}{2}}\Bigg[\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)-\frac{2}{3}\Bigg]+\text{C}$

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