evaluate the integrals\n35. $$ \\int \\frac { e ^ { t } d t } { e ^ { 2 t } + 3 e ^ { t } + 2 } $$

Explanation:

Step1: Substitute (u = e^{t})

Let (u = e^{t}), then (du=e^{t}dt). The integral (\int\frac{e^{t}dt}{e^{2t}+3e^{t}+2}) becomes (\int\frac{du}{u^{2}+3u + 2}).

Step2: Factor the denominator

Factor (u^{2}+3u + 2=(u + 1)(u + 2)). Then, use partial - fraction decomposition. Let (\frac{1}{(u + 1)(u + 2)}=\frac{A}{u+1}+\frac{B}{u + 2}). Cross - multiply: (1=A(u + 2)+B(u + 1)). Set (u=-1), then (1 = A(-1+2)+B(-1 + 1)), so (A = 1). Set (u=-2), then (1=A(-2 + 2)+B(-2+1)), so (B=-1). So (\frac{1}{(u + 1)(u + 2)}=\frac{1}{u + 1}-\frac{1}{u + 2}).

Step3: Integrate the partial - fractions

(\int\frac{du}{u^{2}+3u + 2}=\int\left(\frac{1}{u + 1}-\frac{1}{u + 2}\right)du). Using the formula (\int\frac{1}{x}dx=\ln|x|+C), we have (\int\frac{1}{u + 1}du-\int\frac{1}{u + 2}du=\ln|u + 1|-\ln|u + 2|+C).

Step4: Substitute back (u = e^{t})

Since (u = e^{t}), the integral (\int\frac{e^{t}dt}{e^{2t}+3e^{t}+2}=\ln\left|e^{t}+1\right|-\ln\left|e^{t}+2\right|+C).

Answer:

(\ln\left|e^{t}+1\right|-\ln\left|e^{t}+2\right|+C)