Question

integrate (√{x²+1} (log(x²+1)-2log x))/(x⁴) dx

simply,

$\displaystyle\sf\int\dfrac{\sqrt{x^2+1}(log(x^2+1)-2\:log\:x)}{x^4}\:dx$​

Answer 1

★Given:-

$\leadsto \int \sf \dfrac{(\sqrt{x^2+1} (log(x^2+1)-2log x))}{x^4 }dx$

We need to know:

$\leadsto \sf (nlogm=logm^n)\\\\\leadsto \sf logm-logn=log\frac{m}{n} \\\\\leadsto \sf \int f(x).g(x)dx =F(x).g(x)- \int F(x).g'(x)dx$

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Now,

$:\implies \int \sf \dfrac{(\sqrt{x^2+1} (log(x^2+1)-2log x))}{x^4 }dx\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(x^2+1)-logx^2]\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(\dfrac{x^2+1}{x^2} )]\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(1+\dfrac{1}{x^2})] \\\\\\:\implies \sf \dfrac{1}{x^3} \sqrt{\dfrac{x^2 +1}{x^2} } [log(1+\dfrac{1}{x^2})]\\\\\\:\implies \sf \dfrac{1}{x^3} \sqrt{1+\dfrac{1}{x^2} } [log(1+\dfrac{1}{x^2})]$

Putting 1+1/x² = t,

$:\implies \sf \dfrac{-2}{x^3} dx=dt\\\\\\:\implies \sf I = \int \dfrac{1}{x^3} \sqrt{1+\dfrac{1}{x^2} } log(1+\dfrac{1}{x^2})dx\\\\\\:\implies \sf I =\dfrac{-1}{2} \int \sqrt{t}\ log\ tdt\\\\\\:\implies \sf I =\dfrac{-1}{2} \int t^{1/2}logtdt$

Integrating by parts,

$:\implies \sf I = \dfrac{-1}{2} \left [logt.\int t^{1/2}}dt-\int \left \{ (\dfrac{d}{dt}logt)\int t^{1/2}dt \right \} dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2} \left [logt.\dfrac{t^{3/2}}{3/2} -\int \dfrac{1}{t} .\dfrac{t^{3/2}}{3/2} dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2} \left [\dfrac{2}{3} t^{3/2}logt-\dfrac{2}{3} \int t^{1/2}dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2}\left [\dfrac{2}{3} t^{3/2}logt -\dfrac{4}{9} t^{3/2} \right ]$

$\displaystyle \sf :\implies \dfrac{-1}{3} t^{ 3/2}logt+\dfrac{2}{9} t^{3/2}\\\\\\:\implies \sf \dfrac{-1}{3} t^{3/2}(logt-\frac{2}{3} )$

Now substituting the value of t,

$\mapsto \underline{\boxed{\bold{\frac{-1}{3}(1+\frac{1}{x^2})^{3/2}\left [log(1+\frac{1}{x^2} )-\frac{2}{3}\right ] +C}}}$

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Answer 2

★Given:-

$\leadsto \int \sf \dfrac{(\sqrt{x^2+1} (log(x^2+1)-2log x))}{x^4 }dx$

We need to know:

$\leadsto \sf (nlogm=logm^n)\\\\\leadsto \sf logm-logn=log\frac{m}{n} \\\\\leadsto \sf \int f(x).g(x)dx =F(x).g(x)- \int F(x).g'(x)dx$

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Now,

$:\implies \int \sf \dfrac{(\sqrt{x^2+1} (log(x^2+1)-2log x))}{x^4 }dx\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(x^2+1)-logx^2]\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(\dfrac{x^2+1}{x^2} )]\\\\\\:\implies \sf \dfrac{\sqrt{x^2+1} }{x^4} [log(1+\dfrac{1}{x^2})] \\\\\\:\implies \sf \dfrac{1}{x^3} \sqrt{\dfrac{x^2 +1}{x^2} } [log(1+\dfrac{1}{x^2})]\\\\\\:\implies \sf \dfrac{1}{x^3} \sqrt{1+\dfrac{1}{x^2} } [log(1+\dfrac{1}{x^2})]$

Putting 1+1/x² = t,

$:\implies \sf \dfrac{-2}{x^3} dx=dt\\\\\\:\implies \sf I = \int \dfrac{1}{x^3} \sqrt{1+\dfrac{1}{x^2} } log(1+\dfrac{1}{x^2})dx\\\\\\:\implies \sf I =\dfrac{-1}{2} \int \sqrt{t}\ log\ tdt\\\\\\:\implies \sf I =\dfrac{-1}{2} \int t^{1/2}logtdt$

Integrating by parts,

$:\implies \sf I = \dfrac{-1}{2} \left [logt.\int t^{1/2}}dt-\int \left \{ (\dfrac{d}{dt}logt)\int t^{1/2}dt \right \} dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2} \left [logt.\dfrac{t^{3/2}}{3/2} -\int \dfrac{1}{t} .\dfrac{t^{3/2}}{3/2} dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2} \left [\dfrac{2}{3} t^{3/2}logt-\dfrac{2}{3} \int t^{1/2}dt \right ]\\\\\\:\implies \sf \dfrac{-1}{2}\left [\dfrac{2}{3} t^{3/2}logt -\dfrac{4}{9} t^{3/2} \right ]$

$\displaystyle \sf :\implies \dfrac{-1}{3} t^{ 3/2}logt+\dfrac{2}{9} t^{3/2}\\\\\\:\implies \sf \dfrac{-1}{3} t^{3/2}(logt-\frac{2}{3} )$

Now substituting the value of t,

$\mapsto \underline{\boxed{\bold{\frac{-1}{3}(1+\frac{1}{x^2})^{3/2}\left [log(1+\frac{1}{x^2} )-\frac{2}{3}\right ] +C}}}$

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