Question

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

${\int }_1^4 \frac{2x+3}{x^2+3x+4}dx$

Short answer

Ans: The exact value of,${\int }_1^4 \frac{2x+3}{x^2+3x+4}dx=\ln \left(32\right)-\ln \left(8\right)$

Step-by-step solution

Step 1. Given information.

given,

${\int }_1^4 \frac{2x+3}{x^2+3x+4}dx$

Step 2. The objective is to determine the exact value of the definite integral. 

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_1^4 \frac{2x+3}{x^2+3x+4}dx\\ ={\int }_1^4 \frac{1}{u}du\\ =[\ln (\left|u\right|){]}_1^4\\ =\ln \left|4^2+3\left(4\right)+4\right|-\ln \left|(1{)}^2+3(1)+4\right|\\ =\ln (16+12+4)-\ln (1+3+4)\\ =\ln \left(32\right)-\ln \left(8\right)\end{array}$

Therefore, the exact value is $\ln \left(32\right)-\ln \left(8\right)$

Step 3. Check:

The required graph is,