Question

$\displaystyle \sf \int_{0}^{ ln2 } \frac{ {e}^{x} - {e}^{2x} + {e}^{3x} - {e}^{4x} + {e}^{5x} - {e}^{6x} + {e}^{7x} - {e}^{8x} }{1 + {e}^{x} + {e}^{2x} + {e}^{3x} + {e}^{4x} + {e}^{5x} + {e}^{6x} + {e}^{7x} }$​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \sf \int_{0}^{ ln2 } \frac{ {e}^{x} - {e}^{2x} + {e}^{3x} - {e}^{4x} + {e}^{5x} - {e}^{6x} + {e}^{7x} - {e}^{8x} }{1 + {e}^{x} + {e}^{2x} + {e}^{3x} + {e}^{4x} + {e}^{5x} + {e}^{6x} + {e}^{7x} }dx$

can be rewritten as

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{ ({e}^{x} + {e}^{3x} + {e}^{5x} + {e}^{7x}) - ({e}^{2x} + {e}^{4x} + {e}^{6x} + {e}^{8x} )}{(1 + {e}^{2x} + {e}^{4x} + {e}^{6x}) + ({e}^{x} + {e}^{3x} + {e}^{5x} + {e}^{7x} )}dx$

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{ {e}^{x}(1 + {e}^{2x} + {e}^{4x}+ {e}^{6x}) - {e}^{2x}(1 + {e}^{2x} + {e}^{4x} + {e}^{6x} )}{(1 + {e}^{2x} + {e}^{4x} + {e}^{6x}) + {e}^{x}(1 + {e}^{2x} + {e}^{4x} + {e}^{6x} )}dx$

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{({e}^{x}- {e}^{2x})(1 + {e}^{2x} + {e}^{4x} + {e}^{6x} )}{(1 + {e}^{2x} + {e}^{4x} + {e}^{6x})(1 + {e}^{x})}dx$

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{{e}^{x}- {e}^{2x}}{1 + {e}^{x}}dx$

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{{e}^{x}(1- {e}^{x})}{1 + {e}^{x}}dx$

can be further rewritten as

$\rm \:  =  \: \displaystyle \sf \int_{0}^{ ln2 } \frac{(1- {e}^{x})}{1 + {e}^{x}}{e}^{x} \: dx$

To evaluate this integral, we use method of Substitution

So, Substitute

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

$\red{\rm :\longmapsto\:{e}^{x}dx = dy}$

We know,

In definite integrals, when we substitute, we have to change the limits too.

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = {e}^{x} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf ln2 & \sf 2 \end{array}} \\ \end{gathered}$

So, above integral, can be rewritten as

$\rm \:  =  \: \displaystyle \sf \int_{1}^{ 2 } \frac{1 - y}{1 + y} \: dy$

$\rm \:  =  \: - \: \displaystyle \sf \int_{1}^{ 2 } \frac{y - 1}{y + 1} \: dy$

$\rm \:  =  \: - \: \displaystyle \sf \int_{1}^{ 2 } \frac{y + 1 - 1- 1}{y + 1} \: dy$

$\rm \:  =  \: - \: \displaystyle \sf \int_{1}^{ 2 } \frac{y + 1 - 2}{y + 1} \: dy$

$\rm \:  =  \: - \: \displaystyle \sf \int_{1}^{ 2 } \bigg(\frac{y + 1}{y + 1} - \frac{2}{y + 1} \bigg)\: dy$

$\rm \:  =  \: - \: \displaystyle \sf \int_{1}^{ 2 } \bigg(1 - \frac{2}{y + 1} \bigg)\: dy$

$\rm \:  =  \: - \: \bigg[y - 2log |1 + y| \bigg]_{1}^{ 2 }$

$\rm \:  =  \: - \: \bigg[2 - 2log |1 + 2| - 1 + 2log |1 + 1| \bigg]$

$\rm \:  =  \: - \: \bigg[1 - 2log |3| + 2log |2| \bigg]$

$\rm \:  =  \: 2log3 - 2log2 - 1$

Hence,

$\displaystyle \sf \int_{0}^{ ln2 } \frac{ {e}^{x} - {e}^{2x} + {e}^{3x} - {e}^{4x} + {e}^{5x} - {e}^{6x} + {e}^{7x} - {e}^{8x} }{1 + {e}^{x} + {e}^{2x} + {e}^{3x} + {e}^{4x} + {e}^{5x} + {e}^{6x} + {e}^{7x} } dx \\ \\ \bf \:  =  \: 2log3 - 2log2 - 1$

Answer 2

Step-by-step explanation:

given :

• $\displaystyle \sf \int_{0}^{ ln2 } \frac{ {e}^{x} - {e}^{2x} + {e}^{3x} - {e}^{4x} + {e}^{5x} - {e}^{6x} + {e}^{7x} - {e}^{8x} }{1 + {e}^{x} + {e}^{2x} + {e}^{3x} + {e}</li></ul><p></p><p></p><h2>to find :</h2><p></p><h3>\int_{0}^{ ln2 } \frac{ {e}^{x} - {e}^{2x} + {e}^{3x} - {e}^{4x} + {e}^{5x} - {e}^{6x} + {e}^{7x} - {e}^{8x} }{1 + {e}^{x} + {e}^{2x} + {e}^{3x} + {e}^{4x} + {e}^{5x} + {e}^{6x} + {e}^{7x} }$

  solution :

  • please check the attached file