Mathematics

Solve :
$$\displaystyle\int\dfrac {e^x}{-2(1+e^{-x} )^2} .dx $$


SOLUTION
$$\displaystyle I = \int\dfrac {e^x}{-2(1+e^{-x} )^2}dx $$
$$\displaystyle = \dfrac{-1}{2}\int \dfrac{e^{x}.e^{2x}}{(e^{x}+1)^{2}}dx $$
put $$ e^{x} = t $$
$$ e^{x}dx = dt $$
$$\displaystyle = -\dfrac{1}{2}\int \frac{t^{2}}{(t+1)^{2}}dt $$
$$\displaystyle = -\dfrac{1}{2}\int \frac{[(t+1)^{2}-2t-1]}{(t+1)^{2}}dt $$
$$\displaystyle I = [-\dfrac{1}{2}\int dt]+[\dfrac{1}{2}\int \dfrac{2t+1}{(t+1)^{2}}dt] $$
$$\displaystyle A = \int -\dfrac{1}{2}dt = \dfrac{-t}{2} = \dfrac{-e^{x}}{2} $$
$$ A = \dfrac{-e^{x}}{2} $$
$$(B) \displaystyle \dfrac{1}{2}\int \dfrac{(2t+1)}{(t+1)^{2}}dt $$
$$\displaystyle = \dfrac{1}{2}\int \dfrac{2t+2}{(t+1)^{2}}dt-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt $$
$$\displaystyle \Rightarrow \dfrac{1}{2}\int \dfrac{\dfrac{d}{dx}(t+1)^{2}}{(t+1)^{2}}dt-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt $$
$$\displaystyle \Rightarrow \dfrac{1}{2}log |t+1|^{2}-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt $$
$$\displaystyle \Rightarrow log |P| -\dfrac{1}{2}\int \dfrac{dt}{(t+1)^{2}} $$
$$ (t+1) = p $$
$$ dt = dp $$
$$ \displaystyle \Rightarrow log |p|-\frac{1}{2}\int \frac{dp}{p^{2}} $$
$$ \displaystyle \Rightarrow log |p|-\frac{1}{2}(-\frac{1}{p})+c $$
$$ \displaystyle \Rightarrow log |p|+\frac{1}{2p}+c $$
$$\displaystyle \Rightarrow B = log |t+1|+\frac{1}{2(t+1)}+c $$
$$\boxed{I = A+B = \dfrac{-e^{-x}}{2}+|log|e^{x}+1|+\dfrac{1}{2(e^{x}+1)}+c}$$
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Subjective Medium Published on 17th 09, 2020
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