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Find : ∫ E X ( 2 + E X ) ( 4 + E 2 X ) D X . - Mathematics

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प्रश्न

Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 

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उत्तर

I = \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx\]

\[\text{ Let } e^x = t\]
\[ \Rightarrow e^x dx = dt\]

∴ `I =int (dt)/((2+t)(4+t^2))` 

Let ` 1/((t+2)(t^2+ 4)) = A/(t+2) + (Bt +C) / (t^2 + 4)`

⇒ `1 = A(t^2 + 2) + (Bt +c)(t +2)`

⇒ `1 = (A + B) t^2 + (2B +C) t + (4A + 2C)`

Comparing the coefficients of t2

⇒A + B = 0

⇒ A = - B

Comparing the coefficients of t

2B + C = 0 

⇒ C = -2B

Comparing the constant term

4A + 2C = 1

⇒ -4B - 4B = 1

⇒ B = `(-1)/8`

⇒ ∴ `A =1/8, B = -1/8 and C = 1/4` 

∴ `1/((t+2)(t^2 +4)) = 1/(8(t+2)) + (-t+2)/(8(t^2 + 4))`

∴ `I = 1/8 , int1/(t+2)dt + 1/8 int(2-t)/(t^2+4)dt`

= `1/8 int 1/(t+2)dt + 1/4int(dt)/(t^2+4) -1/8int(tdt)/(t^2+ 4)`

`= 1/8log|t|+ tan^-1  t/2 - 1/16 log |t ^2+ 4| +C`

`= 1/8log |e^x| + 1/8 tan^-1  e^x/2 - 1/16 log |e^(2x) + 4| +C`

 

 

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