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```1C
∫
√tan⁡(𝑥)
𝑑𝑥
1 − 𝑠𝑒𝑛2 (𝑥)
U=tan(x)
du= 𝑠𝑒𝑐 2 (𝑥)𝑑𝑥
𝑑𝑢
= 𝑑𝑥
𝑠𝑒𝑐 2 (𝑥)
√U
1
√U
𝑐𝑜𝑠2 (𝑥)
=∫ 2 ∗ 2( ) 𝑑𝑢
𝑐𝑜𝑠 (𝑥) 𝑠𝑒𝑐 𝑥
=∫ 2 ∗
𝑐𝑜𝑠 (𝑥)
=∫
1
𝑑𝑢
√U
𝑑𝑢
1
3
2𝑢2
=
3
+𝑐
2√𝑡𝑎𝑛3 (𝑥)
=
3
+𝑐
2C
∫(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 𝑑𝑥
u =𝑥 2 − 2𝑥 + 5
dv=⁡𝑒 −𝑥
du=(2x-2)dx
v=−𝑒 −𝑥
= −(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 + ∫(2𝑥 − 2)𝑒 −𝑥 𝑑𝑥
u=2x-2
dv=⁡𝑒 −𝑥
du=2dx
v=−𝑒 −𝑥
= −(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 − (2𝑥 − 2)𝑒 −𝑥 + ∫ 2𝑒 −𝑥 𝑑𝑥
= −𝑥 2 𝑒 −𝑥 + 2𝑥𝑒 −𝑥 − 5𝑒 −𝑥 − 2𝑥𝑒 −𝑥 + 2𝑒 −𝑥 + ∫ 2𝑒 −𝑥 𝑑𝑥
= −𝑥 2 𝑒 −𝑥 − 3𝑒 −𝑥 + 2 ∫ 𝑒 −𝑥 𝑑𝑥
= −𝑥 2 𝑒 −𝑥 − 3𝑒 −𝑥 − 2𝑒 −𝑥 + 𝑐
= ⁡ 𝑥 2 𝑒 −𝑥 − 5𝑒 −𝑥 + 𝑐
3c
∫
𝑥2
√𝑥 2 − 4
𝑑𝑥
Soluci&oacute;n
∫
𝑥2
√𝑥 2 − 4
𝑑𝑥
X=2sec(a)
dx= 2sec(a)tan(a)da
=∫
(2sec(a))2
√(2sec(a))2 − 4
=∫
=⁡∫
∗ 2sec(a)tan(a)da
8𝑠𝑒𝑐 3 (𝑎) tan(𝑎)⁡
√4(sec 2 (𝑎)
− 1)
8𝑠𝑒𝑐 3 (𝑎) tan(𝑎)⁡
√4(tan2 (𝑎)
𝑑𝑎
𝑑𝑎
8𝑠𝑒𝑐 3 (𝑎) tan(𝑎)⁡
=∫
𝑑𝑎
2 tan(𝑎)
= 4 ∫ 𝑠𝑒𝑐 3 (𝑎) 𝑑𝑎
= 4 ∫ sec(𝑎) (𝑠𝑒𝑐 2 (𝑎))𝑑𝑎
= 4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎) + 1)𝑑𝑎
= 4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎))𝑑𝑎 + 4 ∫ sec(𝑎) 𝑑𝑎
Entonces
4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎))𝑑𝑎
1
𝑠𝑒𝑛(𝑎) 2
= 4∫
∗ ⁡(
⁡) 𝑑𝑎
cos⁡(𝑎)
cos⁡(𝑎)
= 4 ∫ 𝑠𝑒𝑛(𝑎) ∗
𝑠𝑒𝑛(𝑎)
𝑑𝑎
𝑐𝑜𝑠 3 (𝑎)
U=sen(a) 𝑑𝑣 =
𝑠𝑒𝑛(𝑎)
𝑐𝑜𝑠 3 (𝑎)
du= cos(a)da 𝑣 =
=
1
2𝑐𝑜𝑠 2 (𝑎)
2𝑠𝑒𝑛(𝑎)
1
−
2
∫
∗ cos(𝑎) 𝑑𝑎
𝑐𝑜𝑠 2 (𝑎)
𝑐𝑜𝑠 2 (𝑎)
=
=
𝑑𝑎
2𝑠𝑒𝑛(𝑎)
− 2 ∫ sec(𝑎) 𝑑𝑎
𝑐𝑜𝑠 2 (𝑎)
2𝑠𝑒𝑛(𝑎)
− 2 ln(sec(𝑎) + tan(𝑎)) + 𝑐
𝑐𝑜𝑠 2 (𝑎)
Entonces
4 ∫ sec(𝑎) 𝑑𝑎
= 4 ln(sec(𝑎) + tan(𝑎)) + 𝑐
Por lo tanto
∫
𝑥2
√𝑥 2 − 4
𝑑𝑥 =
2𝑠𝑒𝑛(𝑎)
− 2 ln(sec(𝑎) + tan(𝑎)) + 4 ln(sec(𝑎) + tan(𝑎)) + 𝑐
𝑐𝑜𝑠 2 (𝑎)
=
2𝑠𝑒𝑛(𝑎)
+ 2 ln(sec(𝑎) + tan(𝑎)) + 𝑐
𝑐𝑜𝑠 2 (𝑎)
Reemplazando a
𝑥√𝑥 2 − 4
𝑥 √𝑥 2 − 4
=
+ 2 ln ( +
)+𝑐
2
2
2
4c∞
0
4
𝑑𝑥
2
−∞ 4 + 𝑥
∫
0
4
𝑑𝑥
2
4
+
𝑥
−𝑛
= lim ∫
𝑛→∞
= lim ∫
𝑛→∞
4
𝑥 2
4 (( ) + 1)
2
= lim ∫
𝑛→∞
1
𝑥 2
((2) + 1)
𝑥
𝑢=
2
𝑑𝑥
𝑑𝑥
2𝑑𝑢 = 𝑑𝑥
= lim (2 ∫
𝑛→∞
1
𝑑𝑢)
(𝑢2 + 1)
= lim (2 arctan(𝑢))
𝑛→∞
𝑥
= lim (2 arctan ( ) 𝑛0𝑙 )
𝑛→∞
2
0
𝑛
= lim (2 arctan ( ) − 2arctan⁡( )
𝑛→∞
2
2
𝑛
= lim − 2arctan⁡( )
𝑛→∞
2
𝑛
= −2 lim arctan ( )
𝑛→∞
2
𝑛
= −2𝑎𝑟𝑐𝑡𝑎𝑛 ( lim )
𝑛→∞ 2
𝜋
= −2 arctan(+∞) = −2 = −𝜋
2
```