special Type Of integral

1. PRESENTATION ABOUT:
SPECIAL
TYPE OF INTEGRAL

2. The problem I would like to present,
𝑑𝑧
∫ 𝑑𝑥
𝑥 + 2 𝑥 − 1
Therefore,
=) 𝑥 − 1 = 𝑧
=)𝑥 − 1 = 𝑧2
=) 𝑥 = 𝑧2 + 1
=)
𝑑
𝑑𝑥
(𝑥 − 1) =
𝑑
𝑑𝑥
. 𝑧2
=) 1− 0 =2𝑧.
TYPE-1:

3. Let,
𝐼 = ∫ 2𝑧.
𝑑𝑧
𝑧2 + 1 + 2 𝑧
= 2∫
𝑑𝑧
𝑧2 + 3
=
𝑑𝑧
𝑧2 + 3
2
=2.
1
𝑎
. tan−1 𝑧
3
+ 𝑐 (solved)

4. The problem I would like to present,
∫ 𝐝𝐱
𝐱 + 𝟏 𝐱 + 𝟓
=)𝒙 − 𝟏 = 𝒛 𝟐
=)𝒅𝒙 = 𝟐𝒛. 𝒅𝒛
Let,
𝑰 = ∫ 𝟐𝒛.
𝒅𝒛
𝒛 𝟐
. 𝒛 𝟐
+ 𝟏 + 𝟓
=∫ 𝟐𝒛.
𝒅𝒛
𝒛 𝒛 𝟐+𝟔
=∫ 𝟐𝒛.
𝒅𝒛
𝒛 𝒛 𝟐+𝟔
= ∫ 𝒅𝒛/ 𝒛 𝟐 + 𝟔
𝟐
= 𝟐 𝒍𝒐𝒈{𝒛 + 𝒛 𝟐 + 𝟔
𝟐
𝑨𝒏𝒔
TYPE 2: