this presentation covers the topic of integration by partial fraction...let me know if u finfd this helpful

2. INTEGRATION BY PARTIAL
FRACTION:

4. • Step 1 Factorize denominator into irreducible factors
•
𝑥2
𝑥2−5𝑥+6
=
𝑑𝑥
(𝑥−3)(𝑥−2)
• CASE 1:If factors are linear put in form
𝐴
𝑥+𝑐
+
𝐵
𝑥+𝑑
and find A
and B
•
𝑑𝑥
𝑥−3 (𝑥−2)
=
𝐴
𝑥−3
ⅆ𝑥 +
𝐵
𝑋−2
ⅆ𝑥

5. • CASE 2: If factors are linear but squared put in form:
•
𝑑𝑥
(𝑥−2)(𝑥+1)2 =
𝐴
𝑥−2
dx+
𝐵
(𝑥+1)
dx+
𝐶
(𝑥+1)2dx

6. •CASE 3: If factors are quadratic put in form:
•
𝑑𝑥
𝑥−1 𝑥2−3𝑥−2
=
𝐴
(𝑥−1)
dx+
𝐵𝑥+𝐶
𝑥2−3𝑥−2
dx

7. •CASE 4: If factors are quadratics but squared put in
form:
•
𝑑𝑥
(𝑥−1)(𝑥2−3𝑥−2)2
•=
𝐴
(𝑥−1)
dx +
𝐵𝑥+𝐶
𝑥2−3𝑥−2
dx +
𝐷𝑥+𝐸
(𝑥2−3𝑥−2)2dx

9. EXAMPLE:
•
3𝑥+1
𝑥2−𝑥−6
ⅆ𝑥
•
3𝑥+1
𝑥2−𝑥−6
=
3𝑥+1
(𝑥+2)(𝑥−3)
• Let
3𝑥+1
(𝑥+2)(𝑥−3)
=
𝐴
(𝑥+2)
+
𝐵
(𝑥−3)
------eq 1
• Multiply by (𝑥 + 2)(𝑥 − 3) on both sides
• 3𝑥 + 1=A(x-3)+B(x-2) -------eq 2
• Factorization:
• 𝑥2 − 𝑥 − 6
• 𝑥2 − 3𝑥 + 2𝑥 − 6
• 𝑥 𝑥 − 3 + 2(𝑥 −
3)

10. • Put x-3 =0
• So x=3
• Put value of x in eq 2
• 3(3)+1=A(3-3)+ B (3+2)
• 10=5B
• B=2
• Put x+2=0
• So x=-2
• Put value of x in eq 2
• 3(-2)+1=A(-2-3)+ B(-2+2)
• -5=A(-5)
• A=1

11. • Put values of A and B in eq 1
•
3𝑥+1
(𝑥+2)(𝑥−3)
=
𝐴
(𝑥+2)
+
𝐵
(𝑥−3)
------eq 1
•
3𝑥+1
(𝑥+2)(𝑥−3)
=
1
𝑥+2
+
2
𝑥−3
• Now apply integration
•
3𝑥+1
(𝑥+2)(𝑥−3)
=
1
𝑥+2
ⅆ𝑥 + 2
1
𝑥−3
ⅆ𝑥
=2ln 𝑥 − 3 + ln 𝑥 + 2 +c
(here c is constant)

12. •THANKS FOR UR PRECIOUS
TIME…