1tried but was not able to do that.
please post full soln. if it doesn't bother u.
\int \frac{1}{2e^{2x}+3e^{x}+1}dx
ans ------> -\frac{1}{2}log\left| e^{-2x}+3e^{-x}+2 \right| + \frac{3}{2}log \left|\frac{e^{-x}+1}{e^{-x}+2} \right| + c
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6 Answers
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1yaar govind bhai is right
put e^x = t
but don differentiate it !!!!
1\int \frac{1}{2 t^2 + 3t + 1} \rightarrow \int \frac{1}{( 2 t + 1 ) ( t + 1 )}
ab to solve karo using part fracn
39This method is called fake substitution...break it by partial fractions and then put back ex.