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# Integrate the function $\frac{\large e^{2x}-1}{\large e^{2x}+1}$

Can you answer this question?

Toolbox:
• (i)Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
• (ii)$\int e^xdx=e^x+c.$
Given $I=\int \frac{e^{2x-1}}{e^2x+1}dx$.

This can as $I=\int\frac{ e^x(e^x-e^{-x})}{e^x(e^x+e^{-x})}dx.$

$I=\int\frac{e^x-e^{-x}}{e^x+e^{-x}}dx.$

Put $e^x+e^{-x}=t\Rightarrow (e^x-e^{-x})dx=dt.$

Now substituting for x and dx we get,

$I=\int\frac{dt}{t}dt.$

On integrating we get,

log|t|+c.

Substituting back for t we get,

$\int\frac{e^{2x}-1}{e^{2x}+1}=log|e^x+e^{-x}|+c.$

Hence $\int\frac{e^{2x}-1}{e^{2x}+1}=log|e^x+e^{-x}|+c.$

answered Jan 29, 2013