Question

Make a substitution to express the integrand as a rational function and then evaluate the integral

\(\int {\frac{{{e^{2x}}}}{{{e^{2x}} + 3{e^x} + 2}}} dx\)

Short answer

The value the of given integral is as follow:

\(\int {\frac{{{e^{2x}}}}{{{e^{2x}}3{e^x} + 2}}} dx\)=-ln \(\left( {{e^x} + 1} \right) + 2\ln \left( {{e^x} + 2} \right) + c\)

Step-by-step solution

Step1:-rewrite the integral and simplify

Let u= \({e^x}\) :-du= \({e^x}dx\)\( \to {e^{ - x}}dx = \frac{1}{u}du\)

Therefore , \(\int {\frac{{{e^{2x}}}}{{{e^{2x}}3{e^x} + 2}}} dx\)=\(\int {\frac{{{u^2}}}{{{u^2} + 3u + 2}}} (\frac{1}{u}du)\)

\( = \int {\frac{{{u^2}}}{{{u^2} + 2u + u + 2}}\left( {\frac{1}{u}} \right)} du = \int {\frac{u}{{u\left( {u + 2} \right) + 1\left( {u + 2} \right)}}} du\)

\( = \int {\frac{u}{{\left( {u + 2} \right)\left( {u + 1} \right)}}} du\)

Step 2:- use partial fraction and find value of A&B

\(\int {\frac{U}{{\left( {U + 2} \right)\left( {U + 1} \right)}}} = \frac{2}{{U + 2}} + \frac{{ - 1}}{{U + 1}}\)

: u = UA + UB + A+ 2B\( \to \)U= (A+B )U+ (A+2B)

Now ,comparing the coefficients on the both side

:-A+B=1 \( \to \)1 A+2B=0\( \to \)2

Subtract equation (1) from equation (2)

A=2B - A-B = 0-1\( \to \)B=-1

Substitute B =-1 in equation (1) ,we get a-1=1\( \to \)A=2

\(\int {\frac{U}{{\left( {U + 2} \right)\left( {U + 1} \right)}}} = \frac{2}{{U + 2}} + \frac{{ - 1}}{{U + 1}}\)

find the integral and substitute u=ex

\(\int {\frac{u}{{(u + 2)(u + 1)}}} du = \int {\frac{2}{{u + 2}}} du - \int {\frac{1}{{u + 1}}} du\) \(\left( {:\int {\frac{{f\left( x \right)}}{{f\left( x \right)}}dx = \ln |f\left( x \right)| + c} } \right)\)

=2LN\(\left( {U + 2} \right) - LN\left( {U + 1} \right) + C\)

=\(2\ln \left( {{e^x} + 2} \right) - \ln \left( {{e^x} + 1} \right) + c\) \(\left( {:u = {e^x}} \right)\)

Therefore, the value of the given integral is given below :-

\(\int {\frac{{{e^{2x}}}}{{{e^{2x}}3{e^x} + 2}}} dx\)=\(2\ln \left( {{e^x} + 2} \right) - \ln \left( {{e^x} + 1} \right) + c\)