How do you find the definite integral for: #(2+3x) / (4+x2) dx# for the intervals #[0, 1]#?

Question

Elijah Abram · Accepted Answer

# arc tan(1/2)+3/2ln(5/4).# Suppose that, #I=int_0^1(2+3x)/(4+x^2)dx.# #:. I=int_0^1 2/(4+x^2)dx+int_0^1(3x)/(4+x^2)dx,# #=2int_0^1 1/(2^2+x^2)dx+3/2int_0^1 (2x)/(4+x^2)dx,# #=2[1/2*arc tan(x/2)]_0^1+3/2int_0^1 {d/dx(4+x^2)}/(4+x^2)dx,# #=[arc tan(1/2)-arc tan(0)]+3/2[ln(4+x^2)]_0^1,# #=arc tan(1/2)-0+3/2[ln(4+1^2)-ln(4+0^2)],# #=arc tan(1/2)+3/2(ln5-ln4),# #=arc tan(1/2)+3/2ln(5/4).#

Cameron Alexander · Answer

undefined To find the definite integral of $ \frac{2 + 3x}{4 + x^2} $ with respect to $ x $ over the interval $[0, 1]$, follow these steps:

1. Split the fraction into partial fractions if possible.
2. Integrate each term separately.
3. Evaluate the definite integral using the Fundamental Theorem of Calculus.

If we split $ \frac{2 + 3x}{4 + x^2} $ into partial fractions, we get:

$$ \frac{2 + 3x}{4 + x^2} = \frac{A}{2 + x} + \frac{B}{2 - x} $$

Solve for $ A $ and $ B $, and integrate each term:

$$ A = \frac{5}{4}, \quad B = \frac{3}{4} $$

The integral of $ \frac{5}{4(2 + x)} $ with respect to $ x $ is $ \frac{5}{4} \ln|2 + x| + C_1 $.
The integral of $ \frac{3}{4(2 - x)} $ with respect to $ x $ is $ -\frac{3}{4} \ln|2 - x| + C_2 $.

Now, evaluate each integral from $ x = 0 $ to $ x = 1 $:

$$ \int_{0}^{1} \frac{2 + 3x}{4 + x^2} \, dx = \left( \frac{5}{4} \ln|2 + x| - \frac{3}{4} \ln|2 - x| \right) \bigg|_{0}^{1} $$

$$ = \left( \frac{5}{4} \ln|3| - \frac{3}{4} \ln|1| \right) - \left( \frac{5}{4} \ln|2| - \frac{3}{4} \ln|2| \right) $$

$$ = \frac{5}{4} \ln\left(\frac{3}{2}\right) - \frac{3}{4} \ln(1) - \frac{5}{4} \ln(2) + \frac{3}{4} \ln(2) $$

$$ = \frac{5}{4} \ln\left(\frac{3}{2}\right) - \frac{5}{4} \ln(2) $$

$$ = \frac{5}{4} \ln\left(\frac{3}{4}\right) $$

Therefore, the definite integral of $ \frac{2 + 3x}{4 + x^2} $ over the interval $[0, 1]$ is $ \frac{5}{4} \ln\left(\frac{3}{4}\right) $.