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Evaluate 
int_(-2)^(e^(2)-3)(x+2)/(x+3)dx. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

Evaluate 2e23x+2x+3dx \int_{-2}^{e^{2}-3} \frac{x+2}{x+3} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

Full solution

Q. Evaluate 2e23x+2x+3dx \int_{-2}^{e^{2}-3} \frac{x+2}{x+3} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
  1. Identify Integral: Identify the integral to be solved.\newlineWe need to evaluate the integral of the function (x+2)/(x+3)(x+2)/(x+3) within the limits of integration from x=2x = -2 to x=e23x = e^{2}-3.
  2. Simplify Integrand: Simplify the integrand if possible.\newlineThe integrand (x+2)/(x+3)(x+2)/(x+3) can be rewritten as 1+(1)/(x+3)1 + (-1)/(x+3) by dividing each term in the numerator by the denominator.
  3. Split Integral: Split the integral into two simpler integrals.\newlineThe integral of the sum of two functions is the sum of their integrals. Therefore, we can write:\newlinex+2x+3dx=1dx+1x+3dx\int\frac{x+2}{x+3}\,dx = \int 1\,dx + \int\frac{-1}{x+3}\,dx
  4. Evaluate First Integral: Evaluate the first integral.\newlineThe integral of 11 with respect to xx is xx. So, (1)dx=x\int(1)\,dx = x.
  5. Evaluate Second Integral: Evaluate the second integral.\newlineThe integral of (1)/(x+3)(-1)/(x+3) with respect to xx is lnx+3-\ln|x+3|. So, ((1)/(x+3))dx=lnx+3\int((-1)/(x+3))\,dx = -\ln|x+3|.
  6. Combine Results: Combine the results of the two integrals. The combined result of the integrals is xlnx+3x - \ln|x+3|.
  7. Apply Limits: Apply the limits of integration.\newlineWe need to evaluate xlnx+3x - \ln|x+3| from x=2x = -2 to x=e23x = e^{2}-3.
  8. Calculate Definite Integral: Calculate the definite integral. \newlineF(x)=xlnx+3F(x) = x - \ln|x+3|\newlineF(e23)=(e23)lne23+3F(e^{2}-3) = (e^{2}-3) - \ln|e^{2}-3+3|\newlineF(2)=(2)ln2+3F(-2) = (-2) - \ln|-2+3|\newlineNow, we subtract F(2)F(-2) from F(e23)F(e^{2}-3):\newlineF(e23)F(2)=(e23)lne2[(2)ln1]F(e^{2}-3) - F(-2) = (e^{2}-3) - \ln|e^{2}| - [(-2) - \ln|1|]
  9. Simplify Expression: Simplify the expression.\newlineSince lne2=2\ln|e^{2}| = 2 and ln1=0\ln|1| = 0, we have:\newline(e23)2[(2)0]=e232+2=e23(e^{2}-3) - 2 - [(-2) - 0] = e^{2}-3 - 2 + 2 = e^{2}-3
  10. Write Final Answer: Write the final answer.\newlineThe final value of the integral is e23e^{2}-3.

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