What is #f(x) = int (x^2+x)/((x+3)(x+4) ) dx# if #f(-2)=2 #?

Question

What is #f(x) = int (x^2+x)/((x+3)(x+4)  )   dx# if #f(-2)=2 #?

Emma Aldridge · Accepted Answer

#f(x)=x+4+12Ln(2)+6Ln(x+3)-12Ln(x+4)#

#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#

=#int (x^2+7x+12-6x-12)/[(x+3)*(x+4)]*dx#

=#int dx-int (6x+12)/[(x+3)*(x+4)]*dx#

=#x-int (6x+12)/[(x+3)*(x+4)]*dx+C#

Now I solved #int (6x+12)/[(x+3)*(x+4)]*dx# integral,

#(6x+12)/[(x+3)*(x+4)]=A/(x+3)+B/(x+4)#

After expanding denominator,

#A*(x+4)+B*(x+3)=6x+12#

Set #x=-4#, #-B=12#, so #B=12#

Set #x=-3#, #A=-6#

Hence,

#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#

=#x-[int -6/(x+3)*dx+12/(x+4)*dx]+C#

=#x-(-6Ln(x+3)+12Ln(x+4))+C#

=#x+6Ln(x+3)-12Ln(x+4)+C#

After using #f(-2)=2# condition,

#-2+6Ln(-2+3)-12Ln(-2+4)+C=2#

#C+6Ln(1)-12Ln(2)-2=2#

#C+0-12Ln(2)-2=2#

#C=4+12Ln2#

Thus,

#f(x)=x+4+12Ln(2)+6Ln(x+3)-12Ln(x+4)#

Scarlett Allison · Answer

undefined To find $f(x) = \int \frac{x^2+x}{(x+3)(x+4)} \, dx$ given $f(-2) = 2$, we first simplify and solve the integral and then use the initial condition to solve for the constant of integration.

Simplify the integrand by performing polynomial long division or partial fraction decomposition. The denominator suggests partial fraction decomposition:
$$
\frac{x^2 + x}{(x+3)(x+4)} = \frac{A}{x+3} + \frac{B}{x+4}.
$$

To find $A$ and $B$, multiply both sides by the common denominator to get rid of the denominators:
$$
x^2 + x = A(x+4) + B(x+3).
$$

Now, we'll solve for $A$ and $B$. Setting $x = -3$ gives us $A$, and setting $x = -4$ gives us $B$:
$$
9 - 3 = A(1) \implies A = 6,
$$
$$
16 - 4 = B(-1) \implies B = -12.
$$

So, the integral simplifies to:
$$
\int \left(\frac{6}{x+3} - \frac{12}{x+4}\right) dx.
$$

Integrate term by term:
$$
\int \frac{6}{x+3} dx - \int \frac{12}{x+4} dx = 6\ln|x+3| - 12\ln|x+4| + C.
$$

Now, use the initial condition $f(-2) = 2$:
$$
6\ln|(-2)+3| - 12\ln|(-2)+4| + C = 2,
$$
$$
6\ln(1) - 12\ln(2) + C = 2,
$$
$$
0 - 12\ln(2) + C = 2,
$$
$$
C = 2 + 12\ln(2).
$$

Therefore, the function $f(x)$ is:
$$
f(x) = 6\ln|x+3| - 12\ln|x+4| + 2 + 12\ln(2).
$$

Ezra Adams · Answer

undefined To find $ f(x) $, integrate the given function:

$$ f(x) = \int \frac{x^2 + x}{(x + 3)(x + 4)} \, dx $$

Given that $ f(-2) = 2 $, we can determine the value of $ f(x) $:

$$ f(x) = \int \frac{x^2 + x}{(x + 3)(x + 4)} \, dx = \ln|x + 3| - \ln|x + 4| + C $$

Using the fact that $ f(-2) = 2 $, we substitute $ x = -2 $ and solve for $ C $:

$$ 2 = \ln|-2 + 3| - \ln|-2 + 4| + C $$
$$ 2 = \ln|1| - \ln|2| + C $$
$$ 2 = 0 - \ln|2| + C $$
$$ 2 = -\ln|2| + C $$
$$ C = 2 + \ln|2| $$

Thus, the function $ f(x) $ is:

$$ f(x) = \ln|x + 3| - \ln|x + 4| + (2 + \ln|2|) $$