Answer:

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Verified answer

Step-by-step explanation:

Let Consider the integral

[tex]\sf \: \displaystyle\int\sf \frac{1}{ \sqrt[3]{x} + \sqrt[4]{x} } \: dx \\ \\ [/tex]

To evaluate this integral, we substitute

[tex]\sf \: \sqrt[12]{x} = y \: \: \sf\implies \: x = {y}^{12} \\ \\ [/tex]

[tex]\sf \: \sf\implies \: dx = 12 \: {y}^{11} \: dy \\ \\ [/tex]

So, above integral can be rewritten as

[tex]\sf \: = \: 12\displaystyle\int\sf \frac{ {y}^{11} }{ {y}^{4} + {y}^{3} } \: dy \\ \\ [/tex]

[tex]\sf \: = \: 12\displaystyle\int\sf \frac{ {y}^{8} }{y + 1} \: dy \\ \\ [/tex]

[tex]\sf \: = \: 12\displaystyle\int\sf \frac{ {y}^{8} + 1 - 1 }{y + 1} \: dy \\ \\ [/tex]

[tex]\sf \: = \: 12\displaystyle\int\sf \frac{(y + 1)( {y}^{7} - {y}^{6} + {y}^{5} + {y}^{4} - {y}^{3} + {y}^{2} - y + 1) - 1 }{y + 1} \: dy \\ \\ [/tex]

[tex]\sf \: = \: 12\displaystyle\int\sf \bigg( {y}^{7} - {y}^{6} + {y}^{5} + {y}^{4} - {y}^{3} + {y}^{2} - y + 1 - \frac{1}{y + 1} \bigg)\: dy \\ \\ [/tex]

[tex]\sf =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 + ln |y + 1| \bigg) + c \\ \\ [/tex]

[tex]\sf =12\bigg(\dfrac{ {(\sqrt[12]{x})}^{8} }{8} - \dfrac{ {(\sqrt[12]{x})}^{7} }{7} + \dfrac{ {(\sqrt[12]{x})}^{6} }{6} - \dfrac{ {(\sqrt[12]{x})}^{5} }{5} + \dfrac{ {(\sqrt[12]{x})}^{4} }{4} - \dfrac{ {(\sqrt[12]{x})}^{3} }{3} + \dfrac{ {(\sqrt[12]{x})}^{2} }{2} - \sqrt[12]{x} + ln |\sqrt[12]{x} + 1| \bigg) + c \\ \\ [/tex]

Now, Consider second integral

[tex]\sf \: \displaystyle\int\sf \frac{In(1 + \sqrt[6]{x}) }{ \sqrt[3]{x} + \sqrt{x} } \: dx\\ \\ [/tex]

To evaluate this integral, we substitute

[tex]\sf \: \sqrt[6]{x} = t \: \: \sf\implies x = {t}^{6} \\ \\ [/tex]

[tex]\sf \: \sf\implies dx = 6{t}^{5} \: dt \\ \\ [/tex]

So, above integral can be rewritten as

[tex]\sf \: = \: \displaystyle\int\sf \frac{ln(1 + t)}{ {t}^{3} + {t}^{2} } \times {6t}^{5} \: dt \\ \\ [/tex]

[tex]\sf \: = \: 6\displaystyle\int\sf \frac{ {t}^{3} \: ln(1 + t)}{t + 1} \: dt \\ \\ [/tex]

Now, again substitute

[tex]\sf \: 1 + t = u \\ \\ [/tex]

[tex]\sf \: t = u - 1 \\ \\ [/tex]

[tex]\sf\implies \sf \:d t = du \\ \\ [/tex]

So, above integral can be rewritten as

[tex]\sf \: = \: 6\displaystyle\int\sf \frac{ {(u - 1)}^{3} \: ln(u)}{u} \: du\\ \\ [/tex]

[tex]\sf \: = \: 6\displaystyle\int\sf \frac{ {(u}^{3} - {3u}^{2} + 3u - 1) \: ln(u)}{u} \: du\\ \\ [/tex]

[tex]\sf \: = \: 6\displaystyle\int\sf\bigg( {(u}^{2} - {3u}^{} + 3 ) \: ln(u) - \dfrac{ln(u)}{u}\bigg) \: du\\ \\ [/tex]

[tex]\sf \: = \: 6\displaystyle\int\sf\bigg( {(u}^{2} - {3u}^{} + 3 ) \: ln(u) \bigg) du - 6 \times \dfrac{ {(ln(u))}^{2} }{2} + d \\ \\ [/tex]

Using integration by parts in first integral, we get

[tex]\sf \: = \: 6ln(u) \bigg(\dfrac{ {u}^{3} }{3} - \dfrac{3 {u}^{2} }{2} + 3u\bigg) - 6\displaystyle\int\sf \dfrac{1}{u} \times \bigg(\dfrac{ {u}^{3} }{3} - \dfrac{3 {u}^{2} }{2} + 3u\bigg)du - 3{(ln(t + 1))}^{2} + d \\ \\ [/tex]

[tex]\sf \: = \: 6ln(t + 1) \bigg(\dfrac{ {(t + 1)}^{3} }{3} - \dfrac{3 {(t + 1)}^{2} }{2} + 3(t + 1)\bigg) - 6\displaystyle\int\sf \bigg(\dfrac{ {u}^{2} }{3} - \dfrac{3 {u}^{} }{2} + 3\bigg)du - 3{(ln(t + 1))}^{2} + d \\ \\ [/tex]

[tex]\sf \: = \: 6ln( \sqrt[6]{x} + 1) \bigg(\dfrac{ {(\sqrt[6]{x} + 1)}^{3} }{3} - \dfrac{3 {(\sqrt[6]{x} + 1)}^{2} }{2} + 3(\sqrt[6]{x} + 1)\bigg) - 6 \bigg(\dfrac{ {(t + 1)}^{3} }{9} - \dfrac{3 {(t + 1)}^{2} }{4} + 3(t + 1)\bigg) - 3{(ln(\sqrt[6]{x}+ 1))}^{2} + d \\ \\ [/tex]

[tex]\sf \: = \: 6ln( \sqrt[6]{x} + 1) \bigg(\dfrac{ {(\sqrt[6]{x} + 1)}^{3} }{3} - \dfrac{3 {(\sqrt[6]{x} + 1)}^{2} }{2} + 3(\sqrt[6]{x} + 1)\bigg) - 6 \bigg(\dfrac{ {(\sqrt[6]{x} + 1)}^{3} }{9} - \dfrac{3 {(\sqrt[6]{x} + 1)}^{2} }{4} + 3(\sqrt[6]{x} + 1)\bigg) - 3{(ln(\sqrt[6]{x}+ 1))}^{2} + d \\ \\ [/tex]

Hence, Solution of above given integral is

[tex]\sf = 12\bigg(\dfrac{ {(\sqrt[12]{x})}^{8} }{8} - \dfrac{ {(\sqrt[12]{x})}^{7} }{7} + \dfrac{ {(\sqrt[12]{x})}^{6} }{6} - \dfrac{ {(\sqrt[12]{x})}^{5} }{5} + \dfrac{ {(\sqrt[12]{x})}^{4} }{4} - \dfrac{ {(\sqrt[12]{x})}^{3} }{3} + \dfrac{ {(\sqrt[12]{x})}^{2} }{2} - \sqrt[12]{x} + ln |\sqrt[12]{x} + 1| \bigg) + \\ \\ \sf \: 6ln( \sqrt[6]{x} + 1) \bigg(\dfrac{ {(\sqrt[6]{x} + 1)}^{3} }{3} - \dfrac{3 {(\sqrt[6]{x} + 1)}^{2} }{2} + 3(\sqrt[6]{x} + 1)\bigg) - 6 \bigg(\dfrac{ {(\sqrt[6]{x} + 1)}^{3} }{9} - \dfrac{3 {(\sqrt[6]{x} + 1)}^{2} }{4} + 3(\sqrt[6]{x} + 1)\bigg) - 3{(ln(\sqrt[6]{x}+ 1))}^{2} + e \\ \\ [/tex]