Question

$\displaystyle \sf \red{\int_{0}^{ \infty} \frac{ \ln \big( \frac{1 + {x}^{11} }{1 + {x}^{3} } \big) }{(1 + {x}^{2} ) \ln(x) } \: dx}$





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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \sf {\int_{0}^{ \infty} \frac{ \ln \big( \frac{1 + {x}^{11} }{1 + {x}^{3} } \big) }{(1 + {x}^{2} ) \ln(x) } \: dx}$

Let assume that

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \infty} \frac{ \ln \big( \frac{1 + {x}^{11} }{1 + {x}^{3} } \big) }{(1 + {x}^{2} ) \ln(x) } \: dx}$

To, evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\:x = tany$

$\rm :\longmapsto\:dx = {sec}^{2}y \: dy$

When we substitute in definite integrals, we have to change the limits too.

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf \infty & \sf \dfrac{\pi}{2} \end{array}} \\ \end{gathered}$

So, above integral can be rewritten as

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \big) }{(1 + {tan}^{2}y) \ln(tany) } \: {sec}^{2}y dy}$

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \big) }{({sec}^{2}y) \ln(tany) } \: {sec}^{2}y dy}$

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \big) }{ \ln(tany) } \: dy} - - - (1)$

We know,

$\boxed{\tt{ \displaystyle \sf {\int_{0}^{a} \: f(x) \: dx \: = \: \displaystyle \sf {\int_{0}^{a} \: f(a - x) \: dx} }}}$

Now, I can be rewritten as

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}( \frac{\pi}{2} - y)}{1 + {tan}^{3}(\frac{\pi}{2} - y)} \big) }{ \ln(tan(\frac{\pi}{2} - y)) } \: dy}$

We know,

$\boxed{\tt{ tan\bigg[\dfrac{\pi}{2} - y\bigg] = coty \: }}$

So, we have

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {cot}^{11}y}{1 + {cot}^{3}y} \big) }{ \ln(coty) } \: dy}$

$\rm :\longmapsto\:I = \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + \frac{1}{ {tan}^{11} y} }{1 + \frac{1}{ {tan}^{3} y} } \big) }{ \ln( {(tany)}^{ - 1} ) } \: dy}$

$\rm :\longmapsto\:I = - \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \times \frac{1}{ {tan}^{8}y } \big) }{ \ln(tany) } \: dy}$

$\rm :\longmapsto\:I = - \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \big) - 8ln(tany)}{ \ln(tany) } \: dy}$

$\rm :\longmapsto\:I = - \displaystyle \sf {\int_{0}^{ \frac{\pi}{2} } \frac{ \ln \big( \frac{1 + {tan}^{11}y}{1 + {tan}^{3}y} \big) }{ \ln(tany) } \: dy} + \displaystyle \sf {\int_{0}^{ \frac{\pi}{2}}}8 \: dy$

$\rm :\longmapsto\:I = - I \: + \: \bigg[8y\bigg]_{0}^{ \dfrac{\pi}{2}}$

$\rm :\longmapsto\:2I = 8 \times \dfrac{\pi}{2}$

$\rm :\longmapsto\:2I = 4\pi$

$\bf\implies \:I = 2\pi$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle \sf {\int_{0}^{ \infty} \frac{ \ln \big( \frac{1 + {x}^{11} }{1 + {x}^{3} } \big) }{(1 + {x}^{2} ) \ln(x) } \: dx} = 2\pi \: }}$