How do you integrate #int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )# using partial fractions?

Question

Evelyn Agard · Accepted Answer

#int(3x^2+10x-5)/((x+1)^2(x-2))dx=-4/(x+1)+3ln|x-2| +C#

First things first:

#(3x^2+10x-5)/((x+1)^2(x-2))=A/(x+1)+B/(x+1)^2+C/(x-2)=#

#(A(x+1)(x-2)+B(x-2)+C(x+1)^2)/((x+1)^2(x-2))=#

#(A(x^2-x-2)+Bx-2B+C(x^2+2x+1))/((x+1)^2(x-2))=#

#(Ax^2-Ax-2A+Bx-2B+Cx^2+2Cx+C)/((x+1)^2(x-2))=#

#((A+C)x^2+(-A+B+2C)x+(-2A-2B+C))/((x+1)^2(x-2))#

Thus, we can now say:

#A+C=3# #-A+B+2C=10# #-2A-2B+C=-5#

This system provides us with:

#A=0, B=4, C=3#

Now let's discuss the integral:

#int(3x^2+10x-5)/((x+1)^2(x-2))dx=int(4/(x+1)^2+3/(x-2))dx=#

#int4/(x+1)^2dx+int3/(x-2)dx=#

#int4(x+1)^(-2)dx+3ln|x-2|=#

#-4(x+1)^(-1)+3ln|x-2|=#

#-4/(x+1)+3ln|x-2| +C#

Samantha Adkison · Answer

undefined To integrate the expression $\frac{{3x^2 + 10x -5}}{{(x+1)^2(x-2)}}$ using partial fractions, first express the fraction as a sum of simpler fractions.

Perform long division if necessary to ensure the numerator's degree is less than the denominator's degree. Then, express the fraction in the form:

$$\frac{{3x^2 + 10x -5}}{{(x+1)^2(x-2)}} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2}$$

Find the values of A, B, and C by multiplying through by the common denominator and comparing coefficients.

After finding the values of A, B, and C, integrate each term separately.

Finally, combine the integrals and simplify if necessary.

Luke Alvarez · Answer

undefined To integrate the given rational function $\frac{3x^2 + 10x - 5}{(x+1)^2(x-2)}$ using partial fractions, we first need to express it as a sum of simpler fractions. The denominator has repeated linear factors, so the partial fraction decomposition will have the form:

$$\frac{3x^2 + 10x - 5}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2}$$

Multiplying both sides by $(x+1)^2(x-2)$ to clear the denominators, we get:

$$3x^2 + 10x - 5 = A(x+1)(x-2) + B(x-2) + C(x+1)^2$$

Expanding the right side and collecting like terms, we get:

$$3x^2 + 10x - 5 = A(x^2 - x - 2) + B(x-2) + C(x^2 + 2x + 1)$$

$$3x^2 + 10x - 5 = (A + C)x^2 + (-A + B + 2C)x + (-2A - 2B + C)$$

Comparing coefficients of $x^2$, $x$, and the constant term, we get the following system of equations:

$$A + C = 3$$
$$-A + B + 2C = 10$$
$$-2A - 2B + C = -5$$

Solving this system of equations, we find $A = 1$, $B = 5$, and $C = 2$. Therefore, the partial fraction decomposition of the given expression is:

$$\frac{3x^2 + 10x - 5}{(x+1)^2(x-2)} = \frac{1}{x+1} + \frac{5}{(x+1)^2} + \frac{2}{x-2}$$

Now, we can integrate each term separately:

$$\int \frac{1}{x+1} \, dx = \ln|x+1|$$
$$\int \frac{5}{(x+1)^2} \, dx = -\frac{5}{x+1}$$
$$\int \frac{2}{x-2} \, dx = 2\ln|x-2|$$

So, the integral of $\frac{3x^2 + 10x - 5}{(x+1)^2(x-2)}$ is:

$$\ln|x+1| - \frac{5}{x+1} + 2\ln|x-2| + C$$