𝐸𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜𝑠 𝑅𝑒𝑠𝑢𝑒𝑙𝑡𝑜𝑠

𝑀é𝑡𝑜𝑑𝑜 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

1. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑥2𝑙𝑛𝑥𝑑𝑥

𝑆𝑜𝑙.

𝐶𝑢𝑎𝑙𝑞𝑢𝑖𝑒𝑟 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑑𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑒𝑠; 𝑒𝑠 𝑟𝑒𝑐𝑜𝑚𝑒𝑛𝑑𝑎𝑏𝑙𝑒 ℎ𝑎𝑐𝑒𝑟 𝑙𝑜 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒:

𝐼 = 𝑥2𝑙𝑛𝑥𝑑𝑥 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑢𝑛 𝑝𝑜𝑙𝑖𝑛ó𝑚𝑖𝑜 𝑐𝑜𝑛 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎

𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢 𝐹ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑃𝑎𝑟𝑎 𝑒𝑠𝑡𝑒 𝑐𝑎𝑠𝑜, 𝑙𝑜 𝑟𝑒𝑐𝑜𝑚𝑒𝑛𝑑𝑎𝑏𝑙𝑒 𝑠𝑖𝑒𝑚𝑝𝑟𝑒𝑠 𝑒𝑠 ℎ𝑎𝑐𝑒𝑟 ∶

𝑢 = 𝑙𝑛𝑥 𝑦 𝑣 = 𝑥2𝑑𝑥

𝑑𝑢 =1

𝑥𝑑𝑥 𝑣 =

𝑥3

3

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠:

𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

3𝑥31

𝑥𝑑𝑥

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

3 𝑥2𝑑𝑥

𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

3

1

3𝑥3 + 𝑐

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

9𝑥3 + 𝑐 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜

𝐼 =1

3𝑥3 𝑙𝑛𝑥 −

1

3𝑥3 + 𝑐 𝑆𝑜𝑙.

𝑆𝑒𝑔𝑢𝑛𝑑𝑜 𝑚é𝑡𝑜𝑑𝑜, 𝑒𝑠 𝑟𝑒𝑎𝑙𝑖𝑧𝑎𝑟 𝑢𝑛𝑎 𝑡𝑎𝑏𝑙𝑎 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑛𝑒𝑟𝑎:

𝑢 𝑑𝑣

𝑙𝑛𝑥 𝑥2 𝐸𝑛 𝑙𝑎 𝑝𝑟𝑖𝑚𝑒𝑟𝑎 𝑐𝑜𝑙ú𝑚𝑛𝑎

𝑠𝑒 𝑑𝑒𝑟𝑖𝑣𝑎

𝐸𝑛 𝑙𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑐𝑜𝑙ú𝑚𝑛𝑎

𝑠𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎

1

𝑥

1

3𝑥3

+

𝐿𝑎𝑠 𝑓𝑙𝑒𝑐𝑎ℎ𝑎𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 𝑖𝑛𝑑𝑖𝑐𝑎; 𝑞𝑢𝑒 𝑠𝑒 𝑑𝑒𝑏𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑟 á𝑚𝑏𝑎𝑠 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑒𝑠

𝐿𝑎𝑠 𝑓𝑙𝑒𝑐ℎ𝑎𝑠 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑒𝑠 𝑖𝑛𝑑𝑖𝑐𝑎𝑛 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙.

𝐸𝑛 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑐𝑎𝑠𝑜 𝑠𝑒𝑟í𝑎:

𝐼 = 𝑥2𝑙𝑛𝑥𝑑𝑥 =1

3𝑥3𝑙𝑛𝑥 −

1

𝑥

1

3𝑥3𝑑𝑥

𝑦 𝑎ñ𝑎𝑑𝑖𝑟 𝑒𝑙 𝑠í𝑚𝑏𝑜𝑙𝑜 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑐𝑜𝑛 𝑠𝑢 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑜 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑑𝑥.

𝑒𝑠 𝑑𝑒𝑐𝑖𝑟: 𝑑𝑥

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

3 𝑥2𝑑𝑥

𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 =1

3𝑥3𝑙𝑛𝑥 −

1

9𝑥3 + 𝑐

𝑆𝑜𝑙.

+

−

2. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: (7 + 𝑥 − 3𝑥2)𝑒−𝑥𝑑𝑥

𝑆𝑜𝑙.

𝐻𝑎𝑐𝑒𝑟: 𝐼 = (7 + 𝑥 − 3𝑥2)𝑒−𝑥𝑑𝑥

𝐹𝑢𝑛𝑐𝑖ó𝑛 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑡𝑒: 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑢𝑛 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑐𝑜𝑛 𝑢𝑛𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑐𝑖𝑎𝑙

𝑃𝑎𝑟𝑎 𝑒𝑠𝑡𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛, 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑟𝑒𝑚𝑜𝑠 𝑒𝑙 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑚é𝑡𝑜𝑑𝑜 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠. 𝑢 𝑑𝑣

7 + 𝑥 − 3𝑥2 𝑒−𝑥

+ 1 − 6𝑥

−6

0

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑒𝑎𝑥𝑑𝑥 =1

𝑎𝑒𝑎𝑥 −𝑒−𝑥

𝑒−𝑥

−𝑒−𝑥

−

+

𝐼 = (7 + 𝑥 − 3𝑥2)𝑒−𝑥𝑑𝑥 = − 7 + 𝑥 − 3𝑥2 𝑒−𝑥 − 1 − 6𝑥 𝑒−𝑥 + 6𝑒−𝑥 + 𝑐

𝐿𝑢𝑒𝑔𝑜, 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠:

𝑆𝑜𝑙.

𝑅𝑒𝑑𝑢𝑐𝑖𝑒𝑛𝑑𝑜 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑐𝑖𝑎𝑙:

𝐼 = 𝑒−𝑥 −7 − 𝑥 + 3𝑥2 − 1 + 6𝑥 + 6 + 𝑐

𝐼 = 3𝑥2 + 5𝑥 − 2 𝑒−𝑥 + 𝑐 𝑆𝑜𝑙.

𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

3. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑠𝑒𝑐2𝑥𝑑𝑥

𝑆𝑜𝑙. 𝑆𝑒𝑎: 𝐼 = 𝑥𝑠𝑒𝑐2𝑥𝑑𝑥

𝑢 𝑑𝑣

𝑥 𝑠𝑒𝑐2𝑥

+ 1

0

𝑡𝑎𝑛𝑥

−ln (𝑐𝑜𝑠𝑥) −

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑠𝑒𝑐2𝑥 𝑑𝑥 = 𝑡𝑎𝑛𝑥

𝑡𝑎𝑛𝑥𝑑𝑥 = −ln (𝑐𝑜𝑠𝑥)

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑠𝑜𝑙. 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠:

𝐼 = 𝑥𝑡𝑎𝑛𝑥 + ln 𝑐𝑜𝑠𝑥 + 𝑐 𝑆𝑜𝑙.

+

4. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑎𝑟𝑐𝑠𝑒𝑛𝑥𝑑𝑥

𝑆𝑜𝑙.

𝑠𝑒𝑎: 𝐼 = 𝑎𝑟𝑐𝑠𝑒𝑛𝑥𝑑𝑥

𝑢 𝑑𝑣

𝑎𝑟𝑐𝑠𝑒𝑛𝑥 1

+ 1

1 − 𝑥2 𝑥

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑒𝑠: 𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 −

𝑥

1 − 𝑥2𝑑𝑥

𝐸𝑛 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑦𝑎 𝑛𝑜 𝑒𝑠 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠; 𝑠𝑖𝑛𝑜 𝑚𝑒𝑑𝑖𝑎𝑛𝑡𝑒 𝑢𝑛 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:

𝑆𝑒𝑎: 𝑧 = 1 − 𝑥2 𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑒𝑠 𝑙𝑜 𝑚𝑖𝑠𝑚𝑜:

𝑧2 = 1 − 𝑥2 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

2𝑧𝑑𝑧 = −2𝑥𝑑𝑥

−𝑧𝑑𝑧 = 𝑥𝑑𝑥

+

−

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛: 𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 − 𝑥

1 − 𝑥2𝑑𝑥

𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 − −𝑧𝑑𝑧

𝑧 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 + 𝑑𝑧

𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 + 𝑧 + 𝑐

𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥 + 1 − 𝑥2 + 𝑐 𝑆𝑜𝑙.

5. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑙𝑛𝑥

𝑥3𝑑𝑥

𝑆𝑒𝑎: 𝐼 = 𝑙𝑛𝑥

𝑥3𝑑𝑥

𝑆𝑜𝑙.

𝐼 = 𝑥−3𝑙𝑛𝑥𝑑𝑥 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠 𝑑𝑜𝑛𝑑𝑒: 𝑢 = 𝑙𝑛𝑥

𝑢 𝑑𝑣

𝑙𝑛𝑥 𝑥−3

+ 1

𝑥

1

−2𝑥−2

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑥𝑛𝑑𝑥 =𝑥𝑛+1

𝑛 + 1

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠: 𝐼 = −

1

2𝑥−2𝑙𝑛𝑥 +

1

2 𝑥−3𝑑𝑥

𝐼 = −1

2

𝑙𝑛𝑥

𝑥2+1

2−1

2𝑥−2 + 𝑐

𝐼 = −1

2

𝑙𝑛𝑥

𝑥2−1

4

1

𝑥2+ 𝑐

𝐼 = −1

2

1

𝑥2𝑙𝑛𝑥 +

1

2+ 𝑐

𝑆𝑜𝑙.

+

−

6. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: ln 𝑥 + 1 + 𝑥2 𝑑𝑥

𝑢 𝑑𝑣

ln (𝑥 + 1 + 𝑥2) 1

+ 1

𝑥 + 1 + 𝑥21 +

1

2 1 + 𝑥22𝑥 𝑥

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠:

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 𝑥( 1 + 𝑥2 + 𝑥)

(𝑥 + 1 + 𝑥2) 1 + 𝑥2𝑑𝑥

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 𝑥

1 + 𝑥2𝑑𝑥 𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:

𝑠𝑒𝑎: 𝑧 = 1 + 𝑥2; 𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑒𝑠 𝑙𝑜 𝑚𝑖𝑠𝑚𝑜

𝑧2 = 1 + 𝑥2 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

2𝑧𝑑𝑧 = 2𝑥𝑑𝑥

𝑧𝑑𝑧 = 𝑥𝑑𝑥

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 𝑧𝑑𝑧

𝑧

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 𝑑𝑧

+

−

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 𝑧 + 𝑐 ; 𝑃𝑒𝑟𝑜: 𝑧 = 1 + 𝑥2

𝐼 = 𝑥𝑙𝑛 𝑥 + 1 + 𝑥2 − 1 + 𝑥2 + 𝑐 𝑆𝑜𝑙.

7. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 cos 𝑙𝑛𝑥 𝑑𝑥

𝑆𝑜𝑙:

𝑆𝑒𝑎: 𝐼 = cos 𝑙𝑛𝑥 𝑑𝑥

𝐴 𝑠𝑖𝑚𝑝𝑙𝑒 𝑣𝑖𝑠𝑡𝑎 𝑠𝑒 𝑣𝑒 𝑞𝑢𝑒 𝑠𝑒 𝑑𝑒𝑏𝑒 ℎ𝑎𝑐𝑒𝑟 𝑢𝑛 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑛 𝑒𝑙 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡𝑜 𝑑𝑒

𝑆𝑒𝑎: 𝑧 = 𝑙𝑛𝑥 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑑𝑧 =1

𝑥𝑑𝑥 𝑝𝑒𝑟𝑜: 𝑥 = 𝑒𝑧

𝑑𝑧 =1

𝑒𝑧𝑑𝑥

𝑑𝑥 = 𝑒𝑧𝑑𝑧

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑦 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑑𝑥

𝐼 = 𝑒𝑧𝑐𝑜𝑠𝑧𝑑𝑧…………(1) 𝑛𝑢𝑒𝑣𝑜 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎

𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚é𝑡𝑟𝑖𝑐𝑎:

𝑝𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑙𝑛𝑏 = 𝑐

𝑏 = 𝑒𝑐

𝑃𝑎𝑟𝑎 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 1 , 𝑠𝑒 𝑎𝑝𝑙𝑖𝑐𝑎 𝑒𝑙 𝑚é𝑡𝑜𝑑𝑜 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠.

𝐼 = 𝑒𝑧𝑐𝑜𝑠𝑧𝑑𝑧…………(1)

𝑢 𝑑𝑣

𝑒𝑧 𝑐𝑜𝑠𝑧

+

− 𝑒𝑧 𝑠𝑒𝑛𝑧

−𝑐𝑜𝑠𝑧 𝑒𝑧

+ 𝑁𝑜𝑡𝑎

E𝑛 𝑐𝑎𝑑𝑎 𝑙í𝑛𝑒𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑑𝑒𝑏𝑒 𝑖𝑟 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑑𝑜 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑠𝑖𝑔𝑛𝑜.

𝐷𝑒 𝑙𝑎 𝑚𝑖𝑠𝑚𝑎 𝑚𝑎𝑛𝑒𝑟𝑎, 𝑒𝑛 𝑐𝑎𝑑𝑎

𝑙í𝑛𝑒𝑎 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑒 𝑑𝑒𝑏𝑒 𝑖𝑟

𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑛𝑑𝑜 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑠𝑖𝑔𝑛𝑜.

𝐸𝑙 𝑠í𝑔𝑛𝑜 ∴ 𝑖𝑛𝑑𝑖𝑐𝑎: 𝑃𝑜𝑟 𝑡𝑎𝑛𝑡𝑜

𝑎 𝑝𝑎𝑟𝑡𝑖𝑟 𝑑𝑒 𝑎ℎ𝑜𝑟𝑎 𝑠𝑒 𝑑𝑖𝑟á:

∴ 𝐼 = 𝑒𝑧𝑠𝑒𝑛𝑧 + 𝑒𝑧𝑐𝑜𝑠𝑧 − 𝐼 + 𝑐

𝐼 + 𝐼 = 𝑒𝑧𝑠𝑒𝑛𝑧 + 𝑒𝑧𝑐𝑜𝑠𝑧 + 𝑐

2𝐼 = 𝑒𝑧𝑠𝑒𝑛𝑧 + 𝑒𝑧𝑐𝑜𝑠𝑧 + 𝑐

2𝐼 = 𝑒𝑧𝑠𝑒𝑛𝑧 + 𝑒𝑧𝑐𝑜𝑠𝑧 + 𝑐

𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 =1

2𝑒𝑧𝑠𝑒𝑛𝑧 +

1

2𝑒𝑧𝑐𝑜𝑠𝑧 + 𝐶 ; 𝐷𝑜𝑛𝑑𝑒: 𝐶 =

𝑐

2

+

𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜: 𝐼 =1

2𝑒𝑧𝑠𝑒𝑛𝑧 +

1

2𝑒𝑧𝑐𝑜𝑠𝑧 + 𝐶 ;

𝐼 =1

2𝑒𝑧 𝑠𝑒𝑛𝑧 + 𝑐𝑜𝑠𝑧 + 𝐶 𝑃𝑒𝑟𝑜: 𝑧 = 𝑙𝑛𝑥

𝑅𝑒𝑡𝑜𝑟𝑛𝑎𝑛𝑑𝑜 𝑎 𝑛𝑢𝑒𝑠𝑡𝑟𝑎𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑒𝑠:

𝐼 =1

2𝑒𝑙𝑛𝑥 𝑠𝑒𝑛 𝑙𝑛𝑥 + cos 𝑙𝑛𝑥 + 𝐶 𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑒𝑙𝑛𝑥 = 𝑥 𝐼 =

1

2𝑥 𝑠𝑒𝑛 𝑙𝑛𝑥 + cos 𝑙𝑛𝑥 + 𝐶 ∴

𝑆𝑜𝑙.

8. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑠𝑒𝑛 𝑙𝑛𝑥 𝑑𝑥 𝑅𝑝𝑡𝑎. : 𝑥

2𝑠𝑒𝑛 𝑙𝑛𝑥 − cos (𝑙𝑛𝑥) + 𝑐

9. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑎𝑟𝑐𝑡𝑔2𝑥𝑑𝑥

𝑆𝑒𝑎: 𝐼 = 𝑥𝑎𝑟𝑐𝑡𝑔2𝑥𝑑𝑥

𝑆𝑜𝑙.

𝐸𝑙 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜 𝑦𝑎 𝑛𝑜 𝑒𝑠 tan 𝑠𝑒𝑛𝑐𝑖𝑙𝑙𝑜 𝑑𝑒 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟, 𝑝𝑜𝑟 𝑙𝑜 𝑞𝑢𝑒 𝑑𝑒𝑏𝑒𝑚𝑜𝑠 𝑡𝑒𝑛𝑒𝑟 𝑚𝑢𝑐ℎ𝑜 𝑐𝑢𝑖𝑑𝑎𝑑𝑜

𝑒𝑛 𝑏𝑢𝑠𝑐𝑎𝑟 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑑𝑒𝑐𝑢𝑎𝑑𝑜:

∴ 𝑠𝑒𝑎: 𝑧 = 𝑎𝑟𝑐𝑡𝑔𝑥 𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑒𝑠 𝑙𝑜 𝑚𝑖𝑠𝑚𝑜

𝑡𝑎𝑛𝑧 = 𝑥 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑠𝑒𝑐2𝑧𝑑𝑧 = 𝑑𝑥

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎: 𝐼 = 𝑥(𝑎𝑟𝑐𝑡𝑎𝑛𝑥)2𝑥𝑑𝑥

𝐼 = 𝑧2𝑡𝑎𝑛𝑧𝑠𝑒𝑐2𝑧𝑑𝑧

𝐼 = 𝑧2𝑠𝑒𝑐𝑧𝑠𝑒𝑐𝑧𝑡𝑎𝑛𝑧𝑑𝑧

𝑆𝑒𝑎: 𝑢 = 𝑧2

𝑑𝑢 = 2𝑧𝑑𝑧

𝑦 𝑣 = 𝑠𝑒𝑐𝑧𝑠𝑒𝑐𝑧𝑡𝑎𝑛𝑧𝑑𝑧 ℎ𝑎𝑐𝑖𝑒𝑛𝑑𝑜: 𝑡 = 𝑠𝑒𝑐𝑧

𝑑𝑡 = 𝑠𝑒𝑐𝑧𝑡𝑎𝑛𝑧𝑑𝑧 𝑣 = 𝑡𝑑𝑡

𝑣 =1

2𝑡2

𝑣 =1

2𝑠𝑒𝑐2𝑧

𝑃𝑒𝑟𝑜: 𝑡 = 𝑠𝑒𝑐𝑧

∴ 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠 𝑒𝑠: 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜:

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 −

1

2 2𝑧𝑠𝑒𝑐2𝑧𝑑𝑧 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝑧𝑠𝑒𝑐2𝑧𝑑𝑧…………(1)

𝑛𝑢𝑒𝑣𝑎𝑚𝑒𝑛𝑡𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠:

𝑆𝑒𝑎: 𝑢 = 𝑧 𝑦 𝑣 = 𝑠𝑒𝑐2𝑧𝑑𝑧

𝑑𝑢 = 𝑑𝑧 𝑣 = 𝑡𝑎𝑛𝑧

𝐸𝑥𝑡𝑟𝑎𝑒𝑚𝑜𝑠 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 1 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟, 𝑑𝑖𝑐𝑖𝑒𝑛𝑑𝑜:

𝑆𝑒𝑎: 𝐼1 = 𝑧𝑠𝑒𝑐2𝑧𝑑𝑧

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠: 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼1 = 𝑧𝑡𝑎𝑛𝑧 − 𝑡𝑎𝑛𝑧𝑑𝑧 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑎. 𝑡𝑎𝑛𝑥𝑑𝑥 = −ln (𝑐𝑜𝑠𝑥)

∴ 𝐼1 = 𝑧𝑡𝑎𝑛𝑧 + ln (𝑐𝑜𝑠𝑧)

𝐿𝑢𝑒𝑔𝑜 𝑠𝑒𝑑𝑒𝑏𝑒 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑟 𝑒𝑙 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 (1)

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝑧𝑠𝑒𝑐2𝑧𝑑𝑧…………(1)

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝐼1

∴ 𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝑧𝑡𝑎𝑛𝑧 + 𝑙𝑛 𝑐𝑜𝑠𝑧 + 𝑐

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝑧𝑡𝑎𝑛𝑧 − 𝑙𝑛(cosz) + 𝑐

𝑅𝑒𝑐𝑜𝑟𝑑𝑒𝑚𝑜𝑠 𝑎ℎ𝑜𝑟𝑎 𝑞𝑢𝑒: 𝑧 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥

𝑡𝑎𝑛𝑧 = 𝑥

𝑡𝑎𝑛𝑧 =𝑥

1

𝑧

𝑥

1

1 + 𝑥2

𝑠𝑒𝑐𝑧 = 1 + 𝑥2

𝑐𝑜𝑠𝑧 =1

1 + 𝑥2

𝑜 𝑡𝑎𝑚𝑏𝑖é𝑛

𝑃𝑜𝑟 ú𝑙𝑡𝑖𝑚𝑜 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑒𝑛 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑡𝑜𝑑𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑒𝑠𝑡á 𝑒𝑛 𝑟𝑒𝑐𝑢𝑎𝑑𝑟𝑜:

𝐼 =1

2𝑧2𝑠𝑒𝑐2𝑧 − 𝑧𝑡𝑎𝑛𝑧 − 𝑙𝑛(cosz) + 𝑐 𝑠𝑒𝑐𝑧 = 1 + 𝑥2

𝑡𝑎𝑛𝑧 = 𝑥

𝑐𝑜𝑠𝑧 =1

1 + 𝑥2

𝐼 =1

2𝑎𝑟𝑐𝑡𝑎𝑛𝑥 2 1 + 𝑥2 − 𝑥𝑎𝑟𝑐𝑡𝑎𝑛𝑥 − 𝑙𝑛

1

1 + 𝑥2+ 𝑐

𝑧 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥

𝑂𝑟𝑑𝑒𝑛𝑎𝑛𝑑𝑜:

𝐼 =1

21 + 𝑥2 𝑎𝑟𝑐𝑡𝑎𝑛𝑥 2 − 𝑥𝑎𝑟𝑐𝑡𝑎𝑛𝑥 +

1

2𝑙𝑛 1 + 𝑥2 + 𝑐

𝑆𝑜𝑙.

10. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥𝑑𝑥

𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒: 𝑆𝑒𝑎:

𝑜 𝑡𝑎𝑚𝑏𝑖é𝑛 𝑠𝑒𝑛𝑧 = 𝑥 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑐𝑜𝑠𝑧𝑑𝑧 = 𝑑𝑥

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑐𝑜𝑠𝑧 = 1 − 𝑠𝑒𝑛2𝑧

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼 = 𝑧2𝑐𝑜𝑠𝑧𝑑𝑧

𝑧 = 𝑎𝑟𝑐𝑠𝑒𝑛𝑥

𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑𝑜 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠.

𝑢 𝑑𝑣

+

−

+

𝑧2 𝑐𝑜𝑠𝑧

2𝑧 𝑠𝑒𝑛𝑧

2 −𝑐𝑜𝑠𝑧

0 −𝑠𝑒𝑛𝑧

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠:

𝐼 = 𝑧2𝑠𝑒𝑛𝑧 + 2𝑧𝑐𝑜𝑠𝑧 − 2𝑠𝑒𝑛𝑧 + 𝑐

𝐼 = 𝑥(𝑎𝑟𝑐𝑠𝑒𝑛𝑥)2 + 2 1 − 𝑥2𝑎𝑟𝑐𝑠𝑒𝑛𝑥 − 2𝑥 + 𝑐

𝑧 = 𝑎𝑟𝑐𝑠𝑒𝑛𝑥

∴

𝑠𝑒𝑛𝑧 = 𝑥 𝑜 𝑡𝑎𝑚𝑏𝑖é𝑛

𝑧

𝑥

1

1 − 𝑥2

𝑐𝑜𝑠𝑧 =1 − 𝑥2

1

𝑐𝑜𝑠𝑧 = 1 − 𝑥2

𝑃𝑒𝑟𝑜:

𝑆𝑜𝑙.

11. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑙𝑛 (𝑙𝑛𝑥)

𝑥𝑑𝑥

𝑆𝑜𝑙.

𝑆𝑒𝑎: 𝐼 = 𝑙𝑛 (𝑙𝑛𝑥)

𝑥𝑑𝑥

𝐴 𝑠𝑖𝑚𝑝𝑙𝑒 𝑣𝑖𝑠𝑡𝑎 𝑠𝑒 ℎ𝑎𝑐𝑒 𝑢𝑛 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑛 𝑒𝑙 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑙𝑎

𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎:

∴ 𝑆𝑒𝑎: 𝑧 = 𝑙𝑛𝑥 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑑𝑧 =1

𝑥𝑑𝑥

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼 = 𝑙𝑛𝑧𝑑𝑧 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑𝑜 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑢 𝑑𝑣

+

𝑙𝑛𝑧 1

1

𝑧 𝑧

+

−

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠:

𝐼 = 𝑧𝑙𝑛𝑧 − 𝑑𝑧

𝐼 = 𝑧𝑙𝑛𝑧 − 𝑧 + 𝑐 𝑃𝑒𝑟𝑜: 𝑧 = 𝑙𝑛𝑥

∴ 𝐼 = 𝑙𝑛𝑥𝑙𝑛 𝑙𝑛𝑥 − 𝑙𝑛𝑥 + 𝑐 𝑆𝑜𝑙.

12. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑙𝑛(𝑥 − 1

𝑥 + 1)𝑑𝑥

𝑠𝑒𝑎: 𝑢 = 𝑙𝑛 (𝑥 − 1

𝑥 + 1) 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑑𝑢 =1

𝑥 − 1𝑥 + 1

𝑥 + 1 − (𝑥 − 1)

(𝑥 + 1)2𝑑𝑥

𝑑𝑢 =𝑥 + 1

𝑥 − 1

2

(𝑥 + 1)2𝑑𝑥 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝑑𝑢 =2

𝑥2 − 1𝑑𝑥

𝑣 = 𝑥𝑑𝑥

𝑣 =1

2𝑥2

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠: 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

∴ 𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1−1

2 𝑥2

2

𝑥2 − 1𝑑𝑥 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1−

𝑥2

𝑥2 − 1𝑑𝑥 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑖𝑚𝑝𝑟𝑜𝑝𝑖𝑎

𝑦𝑎 𝑞𝑢𝑒 𝑡𝑖𝑒𝑛𝑒𝑛 𝑒𝑙 𝑚𝑖𝑠𝑚𝑜 𝑔𝑟𝑎𝑑𝑜

𝑝𝑜𝑟 𝑡𝑎𝑛𝑡𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑠𝑒 ℎ𝑎𝑐𝑒 𝑒𝑠 𝑑𝑖𝑣𝑖𝑑𝑖𝑟 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛ó𝑚𝑖𝑜, 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑠𝑒𝑎 𝑝𝑟𝑜𝑝𝑖𝑎:

𝑥2 𝑥2 − 1

1 −𝑥2 + 1

1

𝐿𝑢𝑒𝑔𝑜: 𝑥2

𝑥2 − 1= 1 +

1

𝑥2 − 1

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙:

𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1− (1 +

1

𝑥2 − 1)𝑑𝑥

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑 𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1− 𝑑𝑥 −

1

𝑥2 − 1𝑑𝑥

1

𝑥2 − 𝑎2𝑑𝑥 =

1

2𝑎ln (𝑥 − 𝑎

𝑥 + 𝑎)

𝐸𝑛 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑐𝑎𝑠𝑜:

𝑎2 = 1 ∴ 𝑎 = 1

𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1− 𝑥 −

1

2ln𝑥 − 1

𝑥 + 1+ 𝑐

𝐼 =1

2𝑥2 ln

𝑥 − 1

𝑥 + 1− 𝑥 −

1

2ln𝑥 − 1

𝑥 + 1+ 𝑐

13. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑥2𝑑𝑥

(𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑒𝑛𝑥)2

𝑠𝑒𝑎: 𝐼 =

𝑥2𝑑𝑥

(𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑒𝑛𝑥)2

𝑆𝑜𝑙.

14. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: (𝑥2 + 1)𝑒𝑥

(𝑥 + 1)2𝑑𝑥

𝑆𝑜𝑙.

𝑆𝑒𝑎: 𝐼 = (𝑥2 + 1)𝑒𝑥

(𝑥 + 1)2𝑑𝑥

𝑃𝑢𝑒𝑠𝑡𝑜 𝑞𝑢𝑒 𝑡𝑖𝑒𝑛𝑒𝑛 𝑒𝑙 𝑚𝑖𝑠𝑚𝑜 𝑔𝑟𝑎𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜

𝑥2 + 1

(𝑥 + 1)2

𝐿𝑜 𝑞𝑢𝑒 𝑠𝑒 ℎ𝑎𝑐𝑒 𝑒𝑠 𝑑𝑖𝑣𝑖𝑑𝑖𝑟 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 ∶

𝑥2 + 1 𝑥2 + 2𝑥 + 1

1 −𝑥2 − 1 − 2𝑥

−2𝑥

𝐿𝑢𝑒𝑔𝑜 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛ó𝑚𝑖𝑜 𝑒𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑒 𝑎:

𝑥2 + 1

(𝑥 + 1)2= 1 −

2𝑥

𝑥2 + 2𝑥 + 1

∴ 𝐼 = 1 −2𝑥

𝑥2 + 2𝑥 + 1𝑒𝑥𝑑𝑥

𝐼 = 𝑒𝑥𝑑𝑥 − 2𝑥𝑒𝑥

(𝑥 + 1)2𝑑𝑥

𝐼 = 𝑒𝑥 − 2𝑥𝑒𝑥

(𝑥 + 1)2𝑑𝑥

𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

∴ 𝑆𝑒𝑎: 𝑢 = 𝑥𝑒𝑥

𝑑𝑢 = 𝑒𝑥 + 𝑥𝑒𝑥 𝑑𝑥

𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑑𝑢 = (𝑥 + 1)𝑒𝑥𝑑𝑥

𝑣 = 2

(𝑥 + 1)2𝑑𝑥

𝑆𝑒𝑎: 𝑧 = 𝑥 + 1

𝑑𝑧 = 𝑑𝑥

𝑣 = 2 𝑑𝑧

𝑧2

𝑣 = 2 𝑧−2𝑑𝑧

∴

𝑣 = 2(𝑧−1

−1)

𝑣 = −2

𝑧

𝑣 = −2

𝑥 + 1

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝑑𝑒 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠 𝑒𝑠: 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 = 𝑒𝑥 − 2𝑥𝑒𝑥

(𝑥 + 1)2𝑑𝑥

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎:

𝐼 = 𝑒𝑥 − (𝑢𝑣 − 𝑣𝑑𝑢)

𝐼 = 𝑒𝑥 − (−2𝑥𝑒𝑥

𝑥 + 1− −

2

𝑥 + 1𝑥 + 1 𝑒𝑥𝑑𝑥) 𝐿𝑢𝑒𝑔𝑜

𝑃𝑒𝑟𝑜: 𝑢 = 𝑥𝑒𝑥

𝑣 = −2

𝑥 + 1

𝑑𝑢 = (𝑥 + 1)𝑒𝑥𝑑𝑥

𝐼 = 𝑒𝑥 +2𝑥𝑒𝑥

𝑥 + 1− 2 𝑒𝑥 𝑑𝑥

𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 = 𝑒𝑥 +2𝑥𝑒𝑥

𝑥 + 1− 2𝑒𝑥 + 𝑐

𝐼 =2𝑥𝑒𝑥

𝑥 + 1− 𝑒𝑥 + 𝑐

𝑆𝑜𝑙.

15. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑒𝑥

(𝑥 + 1)2𝑑𝑥 𝑅𝑝𝑡𝑎. −

𝑥𝑒𝑥

𝑥 + 1+ 𝑒𝑥 + 𝑐

16. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑎𝑟𝑐𝑡𝑔 𝑥2 − 1𝑑𝑥

𝑆𝑜𝑙.

𝑆𝑒𝑎: 𝑧 = 𝑥2 − 1

𝑜 𝑡𝑎𝑚𝑏𝑖é𝑛: 𝑧2 = 𝑥2 − 1

2𝑧𝑑𝑧 = 2𝑥𝑑𝑥

𝐼 = 𝑥𝑎𝑟𝑐𝑡𝑔 𝑥2 − 1𝑑𝑥

𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑧𝑑𝑧 = 𝑥𝑑𝑥

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑛𝑢𝑒𝑠𝑡𝑟𝑜 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎:

𝐼 = 𝑧𝑎𝑟𝑐𝑡𝑎𝑔𝑧𝑑𝑧………… . 1 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑠𝑒𝑎: 𝑢 = 𝑎𝑟𝑐𝑡𝑎𝑔𝑧

𝑑𝑢 =1

1 + 𝑧2𝑑𝑧

𝑣 = 𝑧𝑑𝑧

𝑣 =1

2𝑧2

𝐿𝑢𝑒𝑔𝑜 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 =1

2𝑧2𝑎𝑟𝑐𝑡𝑎𝑔𝑧 −

1

2

𝑧2

1 + 𝑧2𝑑𝑧

𝐷𝑖𝑣𝑖𝑑𝑖𝑒𝑛𝑑𝑜:

𝑧2 𝑧2 + 1

1 −𝑧2 − 1

−1

∴ 𝐼 =1

2𝑧2𝑎𝑟𝑐𝑡𝑎𝑔𝑧 −

1

2 (1 −

1

𝑧2 + 1)𝑑𝑧

𝐼 =1

2𝑧2𝑎𝑟𝑐𝑡𝑎𝑔𝑧 −

1

2 𝑑𝑧 −

1

𝑧2 + 1𝑑𝑧

𝐼 =1

2𝑧2𝑎𝑟𝑐𝑡𝑎𝑔𝑧 −

1

2𝑧 +

1

2𝑎𝑟𝑐𝑡𝑎𝑔𝑧 + 𝑐 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜

𝐼 =1

2𝑧2 + 1 𝑎𝑟𝑐𝑡𝑎𝑔𝑧 −

1

2𝑧 + 𝑐

𝑃𝑒𝑟𝑜: 𝑧2 = 𝑥2 − 1

𝑧 = 𝑥2 − 1

∴ 𝐼 =1

2𝑥2 − 1 + 1 𝑎𝑟𝑐𝑡𝑎𝑔 𝑥2 − 1 −

1

2𝑥2 − 1 + 𝑐

𝐼 =1

2𝑥2𝑎𝑟𝑐𝑡𝑎𝑔 𝑥2 − 1 −

1

2𝑥2 − 1 + 𝑐

𝑆𝑜𝑙.

ó

17. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥

(1 − 𝑥2)32

𝑑𝑥

𝑆𝑜𝑙.

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑆𝑒𝑎: 𝑧 = 𝑎𝑟𝑐𝑠𝑒𝑛𝑥 𝑜 𝑙𝑜 𝑞𝑢𝑒 𝑒𝑠 𝑙𝑜 𝑚𝑖𝑠𝑚𝑜

𝑠𝑒𝑛𝑧 = 𝑥 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑛𝑑𝑜 𝑚 𝑚

𝑐𝑜𝑠𝑧𝑑𝑧 = 𝑑𝑥

𝑧

𝑥

1

1 − 𝑥2

𝑠𝑒𝑛𝑧 = 𝑥

𝑐𝑜𝑠𝑧 = 1 − 𝑥2

𝑡𝑎𝑛𝑧 =𝑥

1 − 𝑥2

∴ 𝐿𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑒 𝑎:

𝐼 = 𝑥𝑎𝑟𝑐𝑠𝑒𝑛𝑥𝑑𝑥

1 − 𝑥2(1 − 𝑥2)

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑦 𝑠𝑢 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙

𝐼 = 𝑧𝑠𝑒𝑛𝑧𝑐𝑜𝑠𝑧𝑑𝑧

𝑐𝑜𝑠𝑧𝑐𝑜𝑠2𝑧 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝐼 = 𝑧𝑠𝑒𝑛𝑧𝑑𝑧

𝑐𝑜𝑠2𝑧 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑆𝑒𝑎: 𝑢 = 𝑧

𝑑𝑢 = 𝑑𝑧

𝑣 = 𝑠𝑒𝑛𝑧

𝑐𝑜𝑠2𝑧𝑑𝑧

𝑡 = 𝑐𝑜𝑠𝑧

𝑑𝑡 = −𝑠𝑒𝑛𝑧𝑑𝑧

−𝑑𝑡 = 𝑠𝑒𝑛𝑧𝑑𝑧

𝑣 = −𝑑𝑡

𝑡2

𝑣 =1

𝑡

𝑣 =1

𝑐𝑜𝑠𝑧

𝐿𝑢𝑒𝑔𝑜 𝑙𝑎 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠 𝑒𝑠:

𝐼 = 𝑧𝑠𝑒𝑛𝑧𝑑𝑧

𝑐𝑜𝑠2𝑧

𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 = 𝑧1

𝑐𝑜𝑠𝑧−

1

𝑐𝑜𝑠𝑧𝑑𝑧

𝐼 = 𝑧1

𝑐𝑜𝑠𝑧− 𝑠𝑒𝑐𝑧𝑑𝑧

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑠𝑒𝑐𝑥𝑑𝑥 = 𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)

𝐼 = 𝑧1

𝑐𝑜𝑠𝑧− 𝑙𝑛 𝑠𝑒𝑐𝑧 + 𝑡𝑎𝑛𝑥 + 𝑐

𝑧

𝑥

1

1 − 𝑥2

𝑠𝑒𝑛𝑧 = 𝑥

𝑐𝑜𝑠𝑧 = 1 − 𝑥2

𝑡𝑎𝑛𝑧 =𝑥

1 − 𝑥2

𝑠𝑒𝑐𝑧 =1

𝑐𝑜𝑠𝑧=

1

1 − 𝑥2

𝐼 =1

1 − 𝑥2𝑎𝑟𝑐𝑠𝑒𝑛𝑥 − 𝑙𝑛

1

1 − 𝑥2+

𝑥

1 − 𝑥2+ 𝑐

𝐼 =1

1 − 𝑥2𝑎𝑟𝑐𝑠𝑒𝑛𝑥 − 𝑙𝑛

𝑥 + 1

1 − 𝑥2+ 𝑐

𝐼 =1

1 − 𝑥2𝑎𝑟𝑐𝑠𝑒𝑛𝑥 + 𝑙𝑛

1 − 𝑥2

𝑥 + 1+ 𝑐

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

−𝑙𝑛𝑎

𝑏= 𝑙𝑛

𝑏

𝑎

𝑆𝑜𝑙.

18. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑎𝑟𝑐𝑡𝑎𝑛𝑥

𝑥2𝑑𝑥

𝑆𝑜𝑙.

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑆𝑒𝑎: 𝑧 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥

𝑥 = 𝑡𝑎𝑛𝑧

𝑑𝑥 = 𝑠𝑒𝑐2𝑧𝑑𝑧

𝐼 = 𝑧𝑠𝑒𝑐2𝑧𝑑𝑧

𝑡𝑎𝑛2𝑧

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜

𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑢 = 𝑧

𝑑𝑢 = 𝑑𝑧

𝑣 = 𝑠𝑒𝑐2𝑧𝑑𝑧

𝑡𝑎𝑛2𝑧

𝑠𝑒𝑎: 𝑡 = 𝑡𝑎𝑛𝑧

𝑑𝑡 = 𝑠𝑒𝑐2𝑧𝑑𝑧

𝑣 = 𝑑𝑡

𝑡2= −

1

𝑡= −

1

𝑡𝑎𝑛𝑧

𝑣 = −1

𝑡𝑎𝑛𝑧

𝐿𝑢𝑒𝑔𝑜 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 = −𝑧

𝑡𝑎𝑛𝑧+

1

𝑡𝑎𝑛𝑧𝑑𝑧

𝐼 = −𝑧

𝑡𝑎𝑛𝑧+

1

𝑡𝑎𝑛𝑧𝑑𝑧

𝐼 = −𝑧

𝑡𝑎𝑛𝑧+ 𝑐𝑜𝑡𝑔𝑥𝑑𝑧

𝐼 = −𝑧

𝑡𝑎𝑛𝑧+ 𝑙 𝑛 𝑠𝑒𝑛𝑧 + 𝑐

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑐𝑜𝑡𝑔𝑥𝑑𝑥 = 𝑙𝑛 (𝑠𝑒𝑛𝑥)

𝐼 = −𝑧𝑐𝑜𝑡𝑔𝑧 + 𝑙 𝑛 𝑠𝑒𝑛𝑧 + 𝑐

𝑧

𝑥 1 + 𝑥2

1

𝑧 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥

𝑥 = 𝑡𝑎𝑛𝑧

𝑑𝑒𝑙 𝑡𝑟𝑎𝑛𝑔𝑢𝑙𝑜 𝑠𝑒 𝑡𝑖𝑒𝑛𝑒:

𝐼 = −1

𝑥𝑎𝑟𝑐𝑡𝑎𝑛𝑥 + 𝑙𝑛

𝑥

1 + 𝑥2+ 𝑐

𝑆𝑜𝑙.

19. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑥

1 − 𝑥2𝑙𝑛 (𝑥 + 1

𝑥 − 1) 𝑑𝑥 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑢 = 𝑙𝑛𝑥 + 1

𝑥 − 1

𝑑𝑢 =𝑥 − 1

𝑥 + 1(𝑥 + 1 − 𝑥 − 1

𝑥 − 1 2 )𝑑𝑥

𝑑𝑢 =2

1 − 𝑥2𝑑𝑥

𝑣 = 𝑥

1 − 𝑥2𝑑𝑥

𝑆𝑒𝑎: 𝑡 = 1 − 𝑥2

𝑡2 = 1 − 𝑥2

2𝑡𝑑𝑡 = −2𝑥𝑑𝑥

−𝑡𝑑𝑡 = 𝑥𝑑𝑥

∴ 𝑣 = −𝑡𝑑𝑡

𝑡

𝑣 = − 𝑑𝑡

𝑣 = −𝑡

𝑣 = − 1 − 𝑥2

𝐿𝑢𝑒𝑔𝑜:

𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 = − 1 − 𝑥2𝑙𝑛𝑥 + 1

𝑥 − 1+

2 1 − 𝑥2

1 − 𝑥2𝑑𝑥

𝐼 = − 1 − 𝑥2𝑙𝑛𝑥 + 1

𝑥 − 1+ 2

1

1 − 𝑥2𝑑𝑥

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑥 =𝑥

𝑥

1 − 𝑥2 =1 − 𝑥2

1 − 𝑥2

1

1 − 𝑥2𝑑𝑥 = 𝑎𝑟𝑐𝑠𝑒𝑛𝑥

𝐼 = − 1 − 𝑥2𝑙𝑛𝑥 + 1

𝑥 − 1+ 2

1

1 − 𝑥2𝑑𝑥

𝐼 = − 1 − 𝑥2𝑙𝑛𝑥 + 1

𝑥 − 1+ 2𝑎𝑟𝑐𝑠𝑒𝑛𝑥 + 𝑐

𝑆𝑜𝑙.

20. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒−2𝑥𝑐𝑜𝑠(𝑒−𝑥)𝑑𝑥

𝑆𝑜𝑙.

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑆𝑒𝑎: 𝑧 = 𝑒−𝑥

𝑑𝑧 = −𝑒−𝑥𝑑𝑥

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼 = 𝑒−𝑥𝑒−𝑥cos (𝑒−𝑥)𝑑𝑥

𝐼 = − 𝑧𝑐𝑜𝑠𝑧𝑑𝑧 −𝑑𝑧 = 𝑒−𝑥𝑑𝑥 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑢 𝑑𝑣 𝑧 𝑐𝑜𝑠𝑧

+

−

1 𝑠𝑒𝑛𝑧

−𝑐𝑜𝑠𝑧 0

+

𝐼 = − 𝑧𝑠𝑒𝑛𝑧 + 𝑐𝑜𝑧 + 𝑐

𝐼 = −𝑧𝑠𝑒𝑛𝑧 − 𝑐𝑜𝑠𝑧 + 𝑐 𝑃𝑒𝑟𝑜: 𝑧 = 𝑒−𝑥

𝐼 = −𝑒−𝑥𝑠𝑒𝑛 𝑒−𝑥 − 𝑐𝑜𝑠 𝑒−𝑥 + 𝑐 𝑆𝑜𝑙.

21. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑒2𝑥𝑐𝑜𝑠(𝑒𝑥) 𝑑𝑥 𝑅𝑝𝑡𝑎. 𝑒𝑥𝑠𝑒𝑛 𝑒𝑥 + 𝑐𝑜𝑠 𝑒𝑥 + 𝑐

22. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 + 1 𝑑𝑥

𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑆𝑒𝑎: 𝑧 = 𝑥 + 1

𝑧2 = 𝑥 + 1

2𝑧𝑑𝑧 = 𝑑𝑥

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼 = 2𝑧𝑎𝑟𝑐𝑡𝑎𝑛𝑧 𝑑𝑧

𝐼 = 2 𝑧𝑎𝑟𝑐𝑡𝑎𝑛𝑧 𝑑𝑧

𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠 𝑞𝑢𝑒 𝑦𝑎 𝑓𝑢𝑒 𝑟𝑒𝑠𝑢𝑒𝑙𝑡𝑎 𝑒𝑛 𝑢𝑛 𝑒𝑗𝑒𝑚𝑝𝑙𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟

𝐼 = 21

2𝑧2𝑎𝑟𝑐𝑡𝑔𝑧 −

1

2

𝑧2

1 + 𝑧2𝑑𝑧

𝑢 𝑑𝑣 𝑎𝑟𝑐𝑡𝑎𝑛𝑧 𝑧

+ 1

1 + 𝑧2

1

2𝑧2

+

−

𝑧2 𝑧2 + 1

1 −𝑧2 − 1

−1

∴ 𝐼 = 21

2𝑧2𝑎𝑟𝑐𝑡𝑎𝑛𝑧 −

1

2 (1 −

1

𝑧2 + 1)𝑑𝑧

𝐼 = 𝑧2𝑎𝑟𝑐𝑡𝑎𝑛𝑧 − 𝑑𝑧 + 1

1 + 𝑧2𝑑𝑧

𝐼 = 𝑧2𝑎𝑟𝑐𝑡𝑎𝑛𝑧 − 𝑧 + 𝑎𝑟𝑐𝑡𝑎𝑛𝑧 + 𝑐 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜

𝐼 = 𝑧2 + 1 𝑎𝑟𝑐𝑡𝑎𝑛𝑧 − 𝑧 + 𝑐

𝑅𝑒𝑡𝑜𝑟𝑛𝑎𝑛𝑑𝑜 𝑎 𝑛𝑢𝑒𝑠𝑡𝑟𝑎𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑒𝑠

𝐼 = 𝑥 + 1 + 1 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 + 1 − 𝑥 + 1 + 𝑐

𝑧 = 𝑥 + 1

𝑧2 = 𝑥 + 1

∴

𝐼 = (𝑥 + 2)𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 + 1 − 𝑥 + 1 + 𝑐 𝑆𝑜𝑙.

𝑙𝑛 𝑥 + 1 + 𝑥 𝑑𝑥 23. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑢 = 𝑙𝑛 ( 𝑥 + 1 + 𝑥)

𝑑𝑢 =1

𝑥 + 1 + 𝑥(1

2 𝑥+

1

2 𝑥 + 1)𝑑𝑥

𝑑𝑢 =1

2

1

𝑥 + 1 + 𝑥(1 + 𝑥 + 𝑥

𝑥 1 + 𝑥)𝑑𝑥 𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜

𝑑𝑢 =1

2

1

𝑥(𝑥 + 1)𝑑𝑥

𝑣 = 𝑑𝑥

𝑣 = 𝑥

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛: 𝐼 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 −1

2

𝑥

𝑥(𝑥 + 1)𝑑𝑥

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑆𝑒𝑎: 𝑧 = 𝑥

𝑧2 = 𝑥

2𝑧𝑑𝑧 = 𝑑𝑥

2𝑧𝑑𝑧 = 𝑑𝑥

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 −1

2 𝑧22𝑧𝑑𝑧

𝑧 𝑧2 + 1 ∴

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 − 𝑧2

𝑧2 + 1𝑑𝑧…………(1)

𝐼1

𝐼1 = 𝑧2

𝑧2 + 1𝑑𝑧

𝑀é𝑡𝑜𝑑𝑜 𝑑𝑒 𝑠𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖ó𝑛 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚é𝑡𝑟𝑖𝑐𝑎

𝐶𝑢𝑎𝑛𝑑𝑜 𝑒𝑙 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑𝑜 𝑡𝑖𝑒𝑛𝑒 𝑙𝑎 𝑓𝑜𝑟𝑚𝑎: 𝑥2 + 𝑎2

𝑠𝑢 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑠: 𝑥 = 𝑎𝑡𝑎𝑛𝜃

𝐸𝑙 𝑐𝑎𝑠𝑜 𝑒𝑠 𝑠𝑖𝑚𝑖𝑙𝑎𝑟: 𝐷𝑜𝑛𝑑𝑒 𝑎2 = 1 𝐸𝑠𝑡𝑜 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑞𝑢𝑒 𝑎 = 1

∴ 𝐸𝑙 𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑠:

𝑧 = 𝑡𝑎𝑛𝜃

𝑑𝑧 = 𝑠𝑒𝑐2𝜃𝑑𝜃

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝐼1:

𝐼1 = 𝑧2

𝑧2 + 1𝑑𝑧

𝑧 = 𝑡𝑎𝑛𝜃

𝑑𝑧 = 𝑠𝑒𝑐2𝜃𝑑𝜃

𝐼1 = 𝑡𝑎𝑛2𝜃

𝑡𝑎𝑛2𝜃 + 1𝑠𝑒𝑐2𝜃𝑑𝜃

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

1 + 𝑡𝑎𝑛2𝜃 = 𝑠𝑒𝑐2𝜃

𝐼1 = 𝑡𝑎𝑛2𝜃

𝑠𝑒𝑐𝜃𝑠𝑒𝑐2𝜃𝑑𝜃

𝐼1 = 𝑡𝑎𝑛2𝜃𝑠𝑒𝑐𝜃𝑑𝜃

𝐼1 = 𝑠𝑒𝑐2𝜃 − 1 𝑠𝑒𝑐𝜃𝑑𝜃

𝐼1 = 𝑠𝑒𝑐3𝜃𝑑𝜃 − 𝑠𝑒𝑐𝜃𝑑𝜃

𝐼1 = 𝑠𝑒𝑐3𝜃𝑑𝜃 − 𝑙𝑛 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 𝑑𝜃

𝑠𝑒𝑐𝑥𝑑𝑥 = 𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)

𝐿𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼2 =1

2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 +

1

2𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)

𝑠𝑒𝑐3𝜃𝑑𝜃 𝑓𝑢𝑒 𝑟𝑒𝑠𝑢𝑒𝑙𝑡𝑎 𝑒𝑛 𝑒𝑙 𝑒𝑗𝑒𝑟𝑐𝑖𝑐𝑖𝑜 25 𝑒𝑛 𝐼2

𝐷𝑜𝑛𝑑𝑒 𝑠𝑢 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑒𝑠:

∴ 𝑠𝑒𝑐3𝜃𝑑𝜃 =1

2𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 +

1

2𝑙𝑛 (𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃)

𝐿𝑢𝑒𝑔𝑜:

𝐼1 = 𝑠𝑒𝑐3𝜃𝑑𝜃 − 𝑙𝑛 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 𝑑𝜃

𝐼1 =1

2𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 +

1

2𝑙 𝑛 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 − 𝑙𝑛 (𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃)

𝐼1 =1

2𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 −

1

2𝑙 𝑛 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃

𝑅𝑒𝑡𝑜𝑟𝑛𝑎𝑛𝑑𝑜 𝑎 𝑛𝑢𝑒𝑠𝑡𝑟𝑎𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑒𝑠

𝑧 = 𝑡𝑎𝑛𝜃

𝜃

𝑧 1 + 𝑧2

1

𝐼1 =1

2𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 −

1

2𝑙 𝑛 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃

𝑠𝑒𝑐𝜃 = 1 + 𝑧2

𝑡𝑎𝑛𝜃 = 𝑧

𝐼1 =1

2𝑧 1 + 𝑧2 −

1

2𝑙𝑛 ( 1 + 𝑧2 + 𝑧)

𝑃𝑜𝑟 ú𝑙𝑡𝑖𝑚𝑜: 𝑧 = 𝑥

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝐼1

𝐼1 =1

2𝑥 1 + 𝑥 −

1

2𝑙𝑛 ( 1 + 𝑥 + 𝑥)

𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝐼1 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 (1)

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 − 𝑧2

𝑧2 + 1𝑑𝑧…………(1)

𝐼1

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 − (1

2𝑥 𝑥 + 1 −

1

2𝑙𝑛 𝑥 + 𝑥 + 1 ) + 𝑐

𝐼 = 𝑥𝑙𝑛 𝑥 + 𝑥 + 1 −1

2𝑥 𝑥 + 1 +

1

2𝑙𝑛 𝑥 + 𝑥 + 1 + 𝑐 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑛𝑑𝑜

𝐼 = 𝑥 +1

2ln 𝑥 + 𝑥 + 1 −

1

2𝑥2 + 𝑥 + 𝑐

𝑆𝑜𝑙.

24. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑠𝑒𝑛2 𝑙𝑛𝑥 𝑑𝑥

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒

𝑠𝑒𝑎: 𝑧 = 𝑙𝑛𝑥

𝑑𝑥 = 𝑒𝑧𝑑𝑧

𝑥 = 𝑒𝑧

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑙𝑛𝑎 = 𝑏

𝑎 = 𝑒𝑏

𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝐼 = 𝑒𝑧𝑠𝑒𝑛2𝑧𝑑𝑧

𝑠𝑒𝑛2𝑥 =1 − 𝑐𝑜𝑠2𝑥

2

𝐼 = 𝑒𝑧1 − 𝑐𝑜𝑠2𝑧

2𝑑𝑧

𝐼 =1

2 𝑒𝑧𝑑𝑧 − 𝑒𝑧𝑐𝑜𝑠2𝑧𝑑𝑧

𝑒𝑎𝑥𝑐𝑜𝑠𝑏𝑥𝑑𝑥 =𝑒𝑎𝑥

𝑎2 + 𝑏2(𝑎𝑐𝑜𝑠𝑏𝑥 + 𝑏𝑠𝑒𝑛𝑏𝑥)

𝐼 =1

2𝑒𝑧 −

𝑒𝑧

5(𝑐𝑜𝑠2𝑧 + 2𝑠𝑒𝑛2𝑧) + 𝑐

𝐼 =1

2𝑒𝑧 1 −

1

5(𝑐𝑜𝑠2𝑧 + 2𝑠𝑒𝑛2𝑧) + 𝑐

𝑃𝑒𝑟𝑜: 𝑧 = 𝑙𝑛𝑥 ;

∴ 𝐼 =1

2𝑥 1 −

1

5(𝑐𝑜𝑠 2𝑙𝑛𝑥 + 2𝑠𝑒𝑛(2𝑙𝑛𝑥)) + 𝑐

𝑥 = 𝑒𝑧

𝑆𝑜𝑙.

25. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 𝑒𝑠𝑒𝑛𝑥𝑐𝑜𝑠4𝑥 − 1

𝑐𝑜𝑠3𝑥𝑑𝑥

𝑆𝑒𝑎: 𝐼 = 𝑒𝑠𝑒𝑛𝑥𝑐𝑜𝑠4𝑥 − 1

𝑐𝑜𝑠3𝑥𝑑𝑥

𝐼 = 𝑒𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥 −

1

𝑐𝑜𝑠3𝑥𝑑𝑥

𝐼1 𝐼2

𝐷𝑜𝑛𝑑𝑒:

𝐼1 = 𝑒𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥

𝐼2 = 𝑠𝑒𝑐3𝑥𝑑𝑥

𝐼 = 𝐼1 − 𝐼2………(1)

𝐶𝑎𝑙𝑐𝑢𝑙𝑜 𝑑𝑒 𝐼1

𝐶𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐼1 = 𝑒𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥

𝑆𝑒𝑎: 𝑧 = 𝑠𝑒𝑛𝑥

𝑑𝑧 = 𝑐𝑜𝑠𝑥𝑑𝑥 𝐼1 = 𝑒

𝑧𝑑𝑧

𝐼1 = 𝑒𝑧

𝐼1 = 𝑒𝑠𝑒𝑛𝑥

𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑎

𝐶𝑎𝑙𝑐𝑢𝑙𝑜 𝑑𝑒 𝐼2

𝐼2 = 𝑠𝑒𝑐3𝑥𝑑𝑥 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑐𝑎𝑠𝑜 𝑐𝑢𝑎𝑛𝑑𝑜 𝑙𝑎 𝑝𝑜𝑡𝑒𝑛𝑐𝑖𝑎 𝑒𝑠 𝑖𝑚𝑝𝑎𝑟

𝐼2 = 𝑠𝑒𝑐𝑥𝑠𝑒𝑐2𝑥𝑑𝑥

𝑠𝑒𝑝𝑎𝑟𝑎𝑟 𝑠𝑒𝑐2𝑥; 𝑃𝑎𝑟𝑎 𝑙𝑢𝑒𝑔𝑜 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑟 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠

𝑆𝑒𝑎: 𝑢 = 𝑠𝑒𝑐𝑥

𝑑𝑢 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥

𝑣 = 𝑠𝑒𝑐2𝑥𝑑𝑥

𝑣 = 𝑡𝑎𝑛𝑥

𝐿𝑢𝑒𝑛𝑔𝑜 𝑒𝑛 𝐼2 𝑠𝑒 𝑡𝑖𝑒𝑛𝑒:

𝐼2 = 𝑠𝑒𝑐𝑥𝑠𝑒𝑐2𝑥𝑑𝑥

𝐼2 = 𝑢𝑣 − 𝑣𝑑𝑢

𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑥𝑑𝑥

𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐𝑥𝑡𝑎𝑛2𝑥𝑑𝑥

𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑

𝑠𝑒𝑐2𝑥 = 1 + 𝑡𝑎𝑛2𝑥

𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2𝑥 − 1

𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐𝑥(𝑠𝑒𝑐2𝑥 − 1)𝑑𝑥

𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐3𝑥𝑑𝑥 + 𝑠𝑒𝑐𝑥𝑑𝑥

𝑠𝑒𝑐𝑥𝑑𝑥 = 𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)

𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − 𝐼2 + 𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥) 𝑆𝑢𝑚𝑎𝑛𝑑𝑜 𝐼2 𝑚𝑚

2𝐼2 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥) 𝐷𝑒𝑠𝑝𝑒𝑗𝑎𝑛𝑑𝑜 𝐼2

𝐼2 =1

2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 +

1

2𝑙𝑛 (𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)

𝑃𝑜𝑟 ú𝑙𝑡𝑖𝑚𝑜 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝐼1 𝑒 𝐼2 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 (1)

𝐼 = 𝐼1 − 𝐼2………(1)

𝐼 = 𝑒𝑠𝑒𝑛𝑥 −1

2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 −

1

2𝑙 𝑛 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 + 𝑐

𝑆𝑜𝑙.