Question

can u solve this integration with steps..​

Answer 1

Solution

We have

$\sf \implies \int \: \dfrac{ {e}^{2x} }{ \sqrt[4]{ {e}^{x} - 1 } } dx$

Now Using Substitution Method

$\sf \implies \: let \: \: {e}^{x} - 1 = {t}^{4}$

$\sf \implies \: {e}^{x} dx = 4 {t}^{3} dt$

we can write as

$\sf \implies \int \: \dfrac{ {e}^{x} \times {e}^{x} }{ \sqrt[4]{ {e}^{x} - 1 } } dx$

$\sf \implies \int \dfrac{e^{x} \times 4 {t}^{3}dt }{( {t}^{4}) {}^{ \frac{1}{4} } } \implies \:\int \dfrac{e^{x} \times 4 {t}^{3} dt}{{t} }$

Now We have to eliminate eˣ We can write

$\sf \implies \: \: {e}^{x} - 1 = {t}^{4}$

$\sf \implies \ {e}^{x} = {t}^{4} + 1$

Put the value

$\sf \implies4\int \dfrac{( {t}^{4} + 1) \times {t}^{3} dt}{{t} }$

$\sf \implies \: 4 \int( {t}^{4} + 1) \times {t}^{2} dt$

$\sf \implies \: 4 \int( {t}^{6} + {t}^{2} )dt$

$\sf \implies \: 4 \bigg \{\dfrac{ {t}^{6 + 1} }{6 + 1} + \dfrac{t {}^{2 + 1} }{2 + 1} \bigg \} + c$

$\sf \implies \: 4 \bigg( \dfrac{ {t}^{7} }{7} + \dfrac{ {t}^{3} }{3} \bigg) + c$

$\sf \implies \:4 \bigg( \dfrac{3 {t}^{7} + 7 {t}^{3} }{21} \bigg) + c$

$\sf \implies \: 4 {t}^{3} \bigg( \dfrac{3t {}^{4} + 7 }{21} \bigg) + c$

Now put the value

$\sf \implies \: \: \: {e}^{x} - 1 = {t}^{4}$

$\sf \implies \: 4( {e}^{x} - 1) {}^{ \frac{3}{4} } \bigg( \dfrac{3( {e}^{x} - 1) + 7 }{21} \bigg) + c$

$\sf \implies \: 4( {e}^{x} - 1)^{ \frac{3}{4} } \bigg( \dfrac{3 {e}^{x} - 3 + 7 }{21} \bigg) + c$

$\sf \implies \: 4( {e}^{x} - 1)^{ \frac{3}{4} } \bigg( \dfrac{3 {e}^{x} + 4}{21} \bigg) + c$

$\sf \implies \: \dfrac{4}{21} ( {e}^{x} - 1) ^{ \frac{3}{4} } (3 {e}^{x} + 4) + c$

Answer

$\sf \implies \: \dfrac{4}{21} ( {e}^{x} - 1) ^{ \frac{3}{4} } (3 {e}^{x} + 4) + c$

Option A is correct