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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 241 of 271
Marks: +1, -0
ex(1+ex)(2+ex)\frac{e^x}{(1+e^x)(2+e^x)}
Solution:  
Let I=∫ex(1+ex)(2+ex)dxI=\int \frac{e^x}{(1+e^x)(2+e^x)}dx
Let ex=te^x = t ⇒exdx=dt\Rightarrow e^x dx = dt
∴I=∫dt(1+t)(2+t)\therefore I=\int \frac{dt}{(1+t)(2+t)}
Now let 1(1+t)(2+t)=A1+t+B2+t\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}
⇒1=A(2+t)+B(1+t)\Rightarrow 1 = A (2 + t) + B (1 + t)
Put t=−1t = -1 ∴1=A(2−1)⇒A=1\therefore 1 = A(2-1) \Rightarrow A = 1
Put t=−2t = -2 ∴1=B(1−2)⇒B=−1\therefore 1 = B(1-2) \Rightarrow B = -1
∴I=∫1(1+t)(2+t)dt\therefore I = \int \frac{1}{(1+t)(2+t)} dt =∫(11+t−12+t)dt= \int \left( \frac{1}{1+t} - \frac{1}{2+t} \right) dt
=log⁡(1+t)−log⁡(2+t)+C= \log(1+t) - \log(2+t) + C
=log⁡(1+ex)−log⁡(2+ex)+C= \log(1+e^x) - \log(2+e^x) + C
=log⁡(1+ex2+ex)+C= \log\left( \frac{1+e^x}{2+e^x} \right) + C
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