Question

➟ Please solve with proper explaination​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \bigg(\dfrac{1}{ \sqrt[3]{x} + \sqrt[4]{x} } + \dfrac{ log(1 + \sqrt[6]{x} ) }{ \sqrt[3]{x} + \sqrt{x} } \bigg) dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int\sf \bigg(\dfrac{1}{ \sqrt[3]{x} + \sqrt[4]{x} } + \dfrac{ log(1 + \sqrt[6]{x} ) }{ \sqrt[3]{x} + \sqrt{x} } \bigg) dx$

$\rm :\longmapsto\:I \: = \displaystyle\int\sf \dfrac{1}{ \sqrt[3]{x} + \sqrt[4]{x} }dx+ \displaystyle\int\sf \dfrac{ log(1 + \sqrt[6]{x} ) }{ \sqrt[3]{x} + \sqrt{x} } dx$

$\rm :\longmapsto\:I = I_1 + I_2$

where,

$\rm :\longmapsto\:I_1 = \displaystyle\int\sf \dfrac{1}{ \sqrt[3]{x} + \sqrt[4]{x} }dx$

and

$\rm :\longmapsto\:I_2 = \displaystyle\int\sf \dfrac{ log(1 + \sqrt[6]{x} ) }{ \sqrt[3]{x} + \sqrt{x} } dx$

Consider,

$\rm :\longmapsto\:I_1 = \displaystyle\int\sf \dfrac{1}{ \sqrt[3]{x} + \sqrt[4]{x} }dx$

We use method of Substitution,

$\red{\rm :\longmapsto\:Put \: x = {y}^{12}}$

$\red{\rm :\longmapsto\:dx = 12 {y}^{11} dy}$

So, on substituting values we have

$\rm :\longmapsto\:I_1 = \displaystyle\int\sf \dfrac{12 {y}^{11} }{ {y}^{4} + {y}^{3} } \: dy$

$\rm :\longmapsto\:I_1 =12 \displaystyle\int\sf \dfrac{{y}^{11} }{ {y}^{3}(y + 1)} \: dy$

$\rm :\longmapsto\:I_1 =12 \displaystyle\int\sf \dfrac{{y}^{8} }{(y + 1)} \: dy$

$\rm :\longmapsto\:I_1 =12 \displaystyle\int\sf \dfrac{{y}^{8} + 1 - 1}{(y + 1)} \: dy$

$\rm :\longmapsto\:I_1 =12 \displaystyle\int\sf \dfrac{{y}^{8} - 1}{(y + 1)} \: dy + 12\displaystyle\int\sf \dfrac{dy}{y + 1}$

$\rm = 12\displaystyle\int\sf ( {y}^{7} - {y}^{6} + {y}^{5} - {y}^{4} + {y}^{3} - {y}^{2} + y - 1) + 12 log(y + 1)$

$\rm \: = 12\bigg(\dfrac{ {y}^{8} }{8} - \dfrac{ {y}^{7} }{7} + \dfrac{ {y}^{6} }{6} - \dfrac{ {y}^{5} }{5} + \dfrac{ {y}^{4} }{4} - \dfrac{ {y}^{3} }{3} + \dfrac{ {y}^{2} }{2} - y\bigg) + 12log(y + 1)$

where,

$\red{\rm :\longmapsto\:y = {\bigg(x \bigg) }^{\dfrac{1}{12} }}$

Now,

Consider,

$\rm :\longmapsto\:I_2 = \displaystyle\int\sf \dfrac{ log(1 + \sqrt[6]{x} ) }{ \sqrt[3]{x} + \sqrt{x} } dx$

$\red{\rm :\longmapsto\:Put \: x = {y}^{6}}$

$\red{\rm :\longmapsto\:dx = 6 {y}^{5} dy}$

So, on substituting the values, we have

$\rm :\longmapsto\:I_2 = \displaystyle\int\sf \dfrac{ log(1 + y) }{ {y}^{2} + {y}^{3}} {6y}^{5} \: dy$

$\rm :\longmapsto\:I_2 = \displaystyle\int\sf \dfrac{ log(1 + y) }{ {y}^{2}(1 + {y})} {6y}^{5} \: dy$

$\rm :\longmapsto\:I_2 =6 \displaystyle\int\sf \dfrac{ log(1 + y) }{(1 + y)} {y}^{3} \: dy$

$\rm :\longmapsto\:I_2 =6 \displaystyle\int\sf \dfrac{ log(1 + y) }{(1 + y)} ({y}^{3} + 1 - 1) \: dy$

$\rm :\longmapsto\:I_2 =6 \displaystyle\int\sf \dfrac{ log(1 + y) }{(1 + y)} ({y}^{3} + 1) \: dy - 6 \displaystyle\int\sf \dfrac{ log(1 + y) }{(1 + y)}dy$

$\rm\: =6 \displaystyle\int\sf log(1 + y)( {y}^{2} - y + 1)\: dy - 6 \frac{ {(y + 1)}^{2} }{2}$

$\rm :\longmapsto\:I_2=6 I_3 - 3 {[log(y + 1)]}^{2}$

Consider,

$\rm :\longmapsto\:I_3 = \displaystyle\int\sf log(1 + y)( {y}^{2} - y + 1)dy$

$\red{\rm :\longmapsto\:Put \: 1 + y = t}$

$\red{\rm :\longmapsto\:dy \: = \: dt}$

$\rm :\longmapsto\:I_3 = \displaystyle\int\sf logt\bigg( {(t - 1)}^{2} - t + 1 + 1\bigg) dt$

$\rm :\longmapsto\:I_3 = \displaystyle\int\sf logt\bigg( {t}^{2} + 1 - 2t - t + 1 + 1\bigg) dt$

$\rm :\longmapsto\:I_3 = \displaystyle\int\sf logt\bigg( {t}^{2} - 3t + 3\bigg) dt$

$\rm = logt\displaystyle\int\sf ( {t}^{2} - 3t + 3)dt - \displaystyle\int\sf \bigg(\dfrac{d}{dt}logt\displaystyle\int\sf ( {t}^{2} - 3t + 3)dt\bigg)dt$

$\rm = logt\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg) - \displaystyle\int\sf \dfrac{1}{t}\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg)dt$

$\rm = logt\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg) - \displaystyle\int\sf \bigg(\dfrac{ {t}^{2} }{3} - \dfrac{3 {t}^{} }{2} + 3\bigg)dt$

$\rm = logt\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg) - \bigg(\dfrac{ {t}^{3} }{9} - \dfrac{3 {t}^{2} }{4} + 3t\bigg)$

$\bf \: I_3 = \rm = logt\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg) - \bigg(\dfrac{ {t}^{3} }{9} - \dfrac{3 {t}^{2} }{4} + 3t\bigg)$

Therefore,

$\rm :\longmapsto\:I_2=6 I_3 - 3 {[log(y + 1)]}^{2}$

$\rm :\longmapsto\:I_2=6 logt\bigg(\dfrac{ {t}^{3} }{3} - \dfrac{3 {t}^{2} }{2} + 3t\bigg) - 6\bigg(\dfrac{ {t}^{3} }{9} - \dfrac{3 {t}^{2} }{4} + 3t\bigg) - 3 {[log(y + 1)]}^{2}$

where,

$\red{\rm :\longmapsto\:t = {\bigg(x \bigg) }^{\dfrac{1}{6} } + 1}$

So,

$\bf :\longmapsto\:I = I_1 + I_2$