# A. Evaluate the definite integral. (Round your answer to three decimal places.) \int_1^3 \frac {x...

## Question:

A.Evaluate the definite integral. (Round your answer to three decimal places.)

{eq}\displaystyle \int_1^3 \frac {x + 1}{x(x^2 + 1)} dx {/eq}

B. Use substitution and partial fractions to find the indefinite integral.

{eq}\displaystyle \int \frac {\sec^2 x}{\tan^2 x + 5 \tan x + 6} dx {/eq}

C. Find the area of the region bounded by the graphs of {eq}y = \frac {97 }{ x^2 + 7x + 12}, y = 0, x = 0, {/eq} and {eq}x = 1 {/eq}.

## Partial Fractions and Substitution:

To evaluate different types of integrals, we must use various integration techniques. In problems where the integrand is a rational function, using partial fraction decomposition (also called partial fraction expansion) is often a way to simplify the work needed to evaluate the integral. Partial fractions involves rewriting the integrand so that it is a sum of simpler functions. Another common technique used in integration is substitution, which is an integration technique that is used to undo a derivative that was evaluated with the chain rule. Often, multiple techniques need to be used together in the same problem.

## Answer and Explanation:

a. We will use partial fraction decomposition to evaluate {eq}\int_1^3 \frac {x + 1}{x(x^2 + 1)} dx {/eq}.
Begin by setting up the partial fraction decomposition of the integrand. The denominator is already factored as a product of a linear factor and an irreducible quadratic factor.

{eq}\frac {x + 1}{x(x^2 + 1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} {/eq}

Then multiply both sides of the equation by the factored denominator and simplify to find the constants.

{eq}x+1=A(x^2+1)+(Bx+C)x\\ \\ x+1=Ax^2+A+Bx^2+Cx\\ \\ x+1 = (A+B)x^2 + Cx +A {/eq}

Comparing coefficients, we see that {eq}A=1 , C=1, B=-1 {/eq}

So the integral can be rewritten as

{eq}\begin{align} \displaystyle \int_1^3 \frac {x + 1}{x(x^2 + 1)} dx &= \int_1^3 (\frac{1}{x}-\frac{x}{x^2+1}+\frac{1}{x^2+1} dx \\ &= \ln|x|-\frac{1}{2}\ln|x^2+1|+\tan^{-1}(x) |_1^3 \\ & = \ln(3)-\frac{1}{2}\ln(10)+\tan^{-1}(3)-\ln(1)+\frac{1}{2}\ln(2)-\tan^{-1}(1) \end{align} {/eq}

Simplifying as much as possible, using logarithm rules, we have that the integral equals

{eq}\boxed{ \tan^{-1}(3)+\frac{1}{2}\ln(\frac{9}{5})-\frac{\pi}{4} \approx 0.758 } {/eq}

b. For this integral, we start by doing substitution

{eq}\displaystyle \int \frac {\sec^2 x}{\tan^2 x + 5 \tan x + 6} dx {/eq}

We will use the substitution {eq}u=\tan(x)\\ du=\sec^2(x) dx {/eq} and so our integral becomes

{eq}\displaystyle \int \frac {\sec^2 x}{\tan^2 x + 5 \tan x + 6} dx = \int \frac{du}{u^2+5u+6} = \int\frac{du}{(u+3)(u+2)} {/eq}

Using partial fraction decomposition, we have

{eq}\frac{1}{(u+3)(u+2)} = \frac{A}{u+3} + \frac{B}{u+2}\\ \\ 1=A(u+2) + B(u+3) {/eq}. Comparing coefficients, we find that {eq}A=-1 , B=1 {/eq} and so

{eq}\begin{align} \displaystyle \int\frac{du}{(u+3)(u+2)} &= \int (-\frac{1}{u+3}+\frac{1}{u+2}) du\\ & = -\ln|u+3| + \ln|u+2| + C\\ & = -\ln|\tan(x)+3| + \ln|\tan(x)+2| + C \end{align} {/eq}

So, we have

{eq}\boxed{ \displaystyle \int \frac {\sec^2 x}{\tan^2 x + 5 \tan x + 6} dx = -\ln|\tan(x)+3| + \ln|\tan(x)+2| + C } {/eq}.

c. To find the area between curves, we must find a definite integral of the top curve minus the bottom curve. In this case, the bottom curve is the x axis.

{eq}\int_0^1(\frac{97}{x^2+7x+12}) dx= \int_0^1(\frac{97}{(x+3)(x+4)} )dx {/eq}

Using partial fraction decomposition again

{eq}\frac{97}{(x+3)(x+4)} = \frac{A}{x+3} + \frac{B}{x+4}\\ \\ 97 = A(x+4) + B(x+3) {/eq}

Solving for our coefficients, we find that {eq}A=97, B=-97 {/eq}

{eq}\begin{align} \displaystyle \int_0^1 \frac{97}{(x+3)(x+4)} dx &= \int_0^1 (\frac{97}{x+3} -\frac{97}{x+4}) dx\\ & = 97\ln|x+3|-97\ln|x+4| |_0^1 \\ &= 97\ln(4)-97\ln(5) -97\ln(3)+97\ln(4) \\ & \approx 6.260 \end{align} {/eq}

Therefore,

{eq}\boxed{ \displaystyle \int_0^1 \frac{97}{(x+3)(x+4)} dx \approx 6.260 } {/eq}