Evaluate the given integral. ∫x−1x+2dx

Question

Evaluate the given integral.  ∫x−1x+2dx

reinosodairyshm · Accepted Answer

Step 1
Use Substitution method to determine the value of integral $\int \frac{x - 1}{x + 2} d x$ .
Step 2
substitute x-1=x+2-3 in the denominator and simplify the rational fraction as below,
$\int \frac{(x + 2) - 3}{x + 2} d x = \int \frac{(x + 2)}{x + 2} - \frac{3}{x + 2} d x$
$= \int 1 - \frac{3}{x + 2} d x$
$= \int 1 d x - \int \frac{3}{x + 2} d x$
Step 3
Substitute x+2=t and dx=dt
$\int 1 d x - \int \frac{3}{x + 2} d x = x - \int \frac{3}{t} d t$
$= x - 3 \ln t + c$
$= x - 3 \ln (x + 2) + c$
hence, the value of integral is $x - 3 \ln (x + 2) + c$ .

maul124uk · Accepted Answer

We have:  ∫x−1x+2dx  The degree of the numerator P (x) is greater than or equal to the degree of the denominator Q (x), so we divide the polynomials.  x−1x+2=1+−3x+2  By integrating the whole part, we get:  ∫1⋅dx=x  Integrating further, we get:  ∫−3x+2dx=−3ln⁡(x+2)  Answer:  x−3ln⁡(x+2)+C

Evaluate the given integral. \int \frac{x-1}{x+2}dx

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