How do you integrate 6x2+7x6(x24)(x+2)dx using partial fractions?

2 Answers
Sep 22, 2017

Given: 6x2+7x6(x24)(x+2)dx

Decompose the integrand into partial fractions:

6x2+7x6(x2)(x+2)2=A1x2+B1(x+2)2+C1x+2

Multiply both sides by (x2)(x+2)2:

6x2+7x6=A(x+2)2+B(x2)+C(x2)(x+2)

Find the value of A by letting x=2:

6(2)2+7(2)6=A((2)+2)2+B(22)+C(22)(2+2)

32=A(4)2

A=2

Find the value of B by letting x=2

6(2)2+7(2)6=2(2+2)2+B(22)+C(22)(2+2)

4=B(22)

B=1

Find the value of C by letting x=0

6(0)2+7(0)6=2(0+2)21(02)+C(02)(0+2)

6=8+24C

C=4

Here is the decomposition:

6x2+7x6(x2)(x+2)2=21x21(x+2)2+41x+2

Therefore, we can write the following equation:

6x2+7x6(x24)(x+2)dx=21x2dx1(x+2)2dx+41x+2dx

The integrals on the right are trival:

6x2+7x6(x24)(x+2)dx=2ln|x2|+1x+2+4ln|x+2|+C

Sep 22, 2017

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

Explanation:

First of all the denominator can be put in a different shape as (x2)(x+2)2. The partial fractions can thus be put as Ax2+Bx(x+2)2+Cx+2

The numerator would thus become A(x+2)2+Bx(x2)+C(x24)

On simplification this adds up to A(x2+4x+4)+Bx22Bx+Cx24C or, # (A+B+C)x^2 + (4A-2B)x +4A-4C.

Now comparing like terms with 6x2+7x6, it would be A+B+C=6, 4A-2B= 7 and4A-4C= -6. Solving these three equations, it would be A=2, B=12 and C=72

The integral thus becomes 2x2dx+12x(x+2)2dx+721x+2dx

Or, 2x2dx+12x+22(x+2)2dx+721x+2dx

0r, 2x2dx+4x+2dx1(x+2)2dx

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