Resolução de exercícios de integrais

Prévia do material em texto

Lista de Exercícios - Cálculo Diferencial e Integral II 
 Prof. Dr. Marcelo Paraná e Monitor Ícaro Viterbre 
 
6) Resolva as integrais indefinidas pelos métodos de integração 
estudados: 
a) ∫ 𝑥 cos(2𝑥) 𝑑𝑥 
b) ∫ 𝑥 (−sen2(𝑥) + 1) 𝑑𝑥 
c) ∫
3𝑥−1
𝑥2−𝑥+1
 𝑑𝑥 
d) ∫
2𝑥−1
𝑥2−3𝑥+2 
 𝑑𝑥 
e) ∫ 3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 
f) ∫
𝑥−1
1+𝑥2
⋅ cos(2𝜋) 𝑑𝑥 
g) ∫ cos3(𝑥) 𝑑𝑥 
h) ∫
𝑒ln (𝑥)
1+𝑥4
 𝑑𝑥 
i) ∫ ln (𝑥 + 3) 𝑑𝑥 
j) ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 
k) ∫ 𝑒ln(𝑥
2) ⋅ 2𝑥 𝑑𝑥 
 
Gabarito 
 
6) 
 
a) 
x ⋅sen(2x)
2
+ 
cos(2𝑥)
4
+ C 
b) 
x2
4
+ 
cos(2x)
8
+ 
x ⋅ sen (2x)
4
+ C 
c) 
3 ⋅ln|x2−x+1|
2
+ 
1
√3
 ⋅
arctg (
2x−1
√3
) + C 
d) ln |
(x−2)3
x−1
| + C 
e) 
−3x⋅ e−2x
2
− 
3⋅e−2x 
4
+ C 
f) 
ln|1+x2|
2
− arctg(x) + C 
g) 
 2 sen3(𝑥)+3 sen(𝑥) cos2(𝑥) 
3
+ C 
h) 
arctg (x2)
2
+ C 
i) (x + 3) ⋅ ln(x + 3) − x + C 
j) 
e2x
13
 ⋅ [3 ⋅ sen(3x) + 2 ⋅
cos(3x)] + C 
k) 
x4
2
+ C
 
 
 
a) ∫ 𝑥 ⋅ cos(2𝑥) 𝑑𝑥 
 
𝑢 = 𝑥 𝑑𝑢 = 1 𝑑𝑥 
𝑑𝑣 = cos(2𝑥) 𝑑𝑥 𝑣 = ∫ 𝑑𝑣 
∫ cos(2𝑥) 𝑑𝑥 
𝑤 = 2𝑥 𝑑𝑤 = 2 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
2
 
∫ cos(𝑤) 
𝑑𝑤
2
 
sen(𝑤)
2
 
sen(2𝑥)
2
 
𝑣 =
sen(2𝑥)
2
 
 
∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢 
 
∫ 𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
− ∫
sen(2𝑥)
2
 𝑑𝑥 
 
∫ 𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
−
1
2
⋅ ∫ sen(2𝑥) 𝑑𝑥 
 
∫ sen(2𝑥) 𝑑𝑥 
ℎ = 2𝑥 𝑑ℎ = 2 𝑑𝑥 𝑑𝑥 =
𝑑ℎ
2
 
∫ sen(ℎ)
𝑑ℎ
2
 
−
cos(ℎ)
2
 
−
cos(2𝑥)
2
 
∫ 𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
−
1
2
⋅ (
− cos(2𝑥)
2
) 
𝑥 ⋅ sen(2𝑥)
2
+
cos(2𝑥)
4
+ 𝐶 
 
 
 
b) ∫ 𝑥 (− sen2(𝑥) + 1) 𝑑𝑥 
 
cos2(𝑥) + sen2(𝑥) = 1 cos2(𝑥) = − sen2(𝑥) + 1 
∫ 𝑥 cos2(𝑥) 𝑑𝑥 
𝑢 = 𝑥 𝑑𝑢 = 1 𝑑𝑥 
𝑑𝑣 = cos2(𝑥) 𝑑𝑥 𝑣 = ∫ 𝑑𝑣 
 
∫ cos2(𝑥)𝑑𝑥 
cos2(𝑥) =
1 + cos(2𝑥)
2
 
∫
1 + cos(2𝑥)
2
 𝑑𝑥 
∫
1
2
𝑑𝑥 + ∫
cos(2𝑥)
2
 𝑑𝑥 
 
∫
1
2
𝑑𝑥 =
𝑥
2
 
∫
cos(2𝑥)
2
 𝑑𝑥 
𝑤 = 2𝑥 𝑑𝑤 = 2 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
2
 
∫
cos(𝑤)
2
𝑑𝑤
2
 
sen(𝑤)
4
 
sen(2𝑥)
4
 
𝑣 =
𝑥
2
+
sen(2𝑥)
4
 
 
∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢 
 
∫ 𝑥 (− sen2(𝑥) + 1) 𝑑𝑥 = 𝑥 ⋅ (
𝑥
2
+
sen(2𝑥)
4
) − ∫ (
sen(2𝑥)
4
+
𝑥
2
) 𝑑𝑥 
∫ 𝑥 (− sen2(𝑥) + 1) 𝑑𝑥 = 𝑥 ⋅ (
𝑥
2
+
sen(2𝑥)
4
) −
1
4
⋅ ∫ sen(2𝑥) 𝑑𝑥 − ∫
𝑥
2
 𝑑𝑥 
 
 
 
∫
𝑥
2
 𝑑𝑥 
𝑥2
4
 
 
∫ sen(2𝑥) 𝑑𝑥 
ℎ = 2𝑥 𝑑ℎ = 2 𝑑𝑥 𝑑𝑥 =
𝑑ℎ
2
 
∫ sen(ℎ)
𝑑ℎ
2
 
− cos(ℎ)
2
 
− cos(2𝑥)
2
 
 
∫ 𝑥 (− sen2(𝑥) + 1) 𝑑𝑥 = 𝑥 ⋅ (
𝑥
2
+
sen(2𝑥)
4
) −
1
4
⋅ (
− cos(2𝑥)
2
) −
𝑥2
4
 
 
∫ 𝑥 (− sen2(𝑥) + 1) 𝑑𝑥 =
𝑥2
2
−
𝑥2
4
+
𝑥 ⋅ sen(2𝑥)
4
+
cos(2𝑥)
8
+ 𝐶 
 
𝑥2
2
−
𝑥2
4
=
𝑥2
4
 
 
𝑥2
4
+
cos(2𝑥)
8
+
𝑥 ⋅ sen(2𝑥)
4
+ 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
 
c) ∫
3𝑥−1
𝑥2−𝑥+1
 𝑑𝑥 
 
∆ = −3 
∆∉ ℝ 
𝑥2 − 𝑥 + 1 
𝑥2 − 2 (𝑥) − (
1
2
) + (
1
2
)
2
+
3
4
 
(𝑥 −
1
2
)
2
+ (
√3
2
)
2
 
𝑢2 = (𝑥 −
1
2
)
2
 
𝑎 = (
√3
2
)
2
 
𝑢 = 𝑥 −
1
2
 𝑥 = 𝑢 +
1
2
 𝑑𝑢 = 𝑑𝑥 
 
∫
3𝑥 − 1
(𝑥 −
1
2)
2
+ (
√3
2 )
2 𝑑𝑥 = ∫
3 (𝑢 +
1
2) − 1
𝑢2 + 𝑎2
 𝑑𝑢 
∫
3𝑢 +
3
2 − 1
𝑢2 + 𝑎2
 𝑑𝑢 = ∫
3𝑢 +
1
2
𝑢2 + 𝑎2
 𝑑𝑢 = ∫
3𝑢
𝑢2 + 𝑎2
 𝑑𝑢 + ∫
1
2
𝑢2 + 𝑎2
 𝑑𝑢 
 
∫
3𝑢
𝑢2 + 𝑎2
 𝑑𝑢 = 3 ⋅ ∫
𝑢
𝑢2 + 𝑎2
 𝑑𝑢 
 
𝑤 = 𝑢2 + 𝑎2 𝑑𝑤 = 2𝑢 𝑑𝑢 𝑑𝑢 =
𝑑𝑤
2𝑢
 
3 ⋅ ∫
𝑢
𝑤
𝑑𝑤
2𝑢
=
3
2
⋅ ∫
𝑑𝑤
𝑤
 
3
2
⋅ ln|𝑤| =
3
2
⋅ ln|𝑢2 + 𝑎2| =
3
2
⋅ ln |(𝑥 −
1
2
)
2
+ (
√3
2
)
2
| =
3 ln|𝑥2 − 𝑥 + 1|
2
 
3 ln|𝑥2 − 𝑥 + 1|
2
 
 
 
 
 
∫
1
2
𝑢2 + 𝑎2
 𝑑𝑢 =
1
2
⋅ ∫
1
𝑢2 + 𝑎2
 𝑑𝑢 =
1
2
⋅
1
𝑎
⋅ arctg (
𝑢
𝑎
) =
1
2
⋅
1
√3
2
⋅ arctg (
𝑥 −
1
2
√3
2
) 
 
1
2
⋅
1
√3
2
⋅ arctg (
𝑥 −
1
2
√3
2
) =
1
√3
⋅ arctg (
2𝑥 − 1
√3
) 
1
√3
⋅ arctg (
2𝑥 − 1
√3
) 
 
∫
3𝑥 − 1
𝑥2 − 𝑥 + 1
 𝑑𝑥 =
3 ln|𝑥2 − 𝑥 + 1|
2
+
1
√3
⋅ arctg (
2𝑥 − 1
√3
) 
 
3 ln|𝑥2 − 𝑥 + 1|
2
+
1
√3
⋅ arctg (
2𝑥 − 1
√3
) + 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
d) ∫
2𝑥−1
𝑥2−3𝑥+2
 𝑑𝑥 
 
Δ = 𝑏2 − 4𝑎𝑐 𝑥 =
−𝑏±√Δ 
2𝑎
 
Δ = (−3)2 − 4 ⋅ 1 ⋅ 2 𝑥 =
−(−3)±√1
2
 
Δ = 1 𝑥 =
3±1
2
 𝑥′ = 1 𝑒 𝑥" = 2 
 
 
2𝑥 − 1
(𝑥 − 2)(𝑥 − 1)
=
𝐴
𝑥 − 2
+
𝐵
𝑥 − 1
 
 
𝑥 = 2 
(2𝑥 − 1)(𝑥 − 2)
(𝑥 − 2)(𝑥 − 1)
=
𝐴(𝑥 − 2)
𝑥 − 2
+
𝐵(𝑥 − 2)
𝑥 − 1
 
 
2𝑥 − 1
𝑥 − 1
= 𝐴 +
𝐵(𝑥 − 2)
𝑥 − 1
 
(2 ⋅ 2) − 1
2 − 1
= 𝐴 +
𝐵(2 − 2)
2 − 1
 
𝐴 = 3 
 
𝑥 = 1 
 
(2𝑥 − 1)(𝑥 − 1)
(𝑥 − 2)(𝑥 − 1)
=
𝐴(𝑥 − 1)
𝑥 − 2
+
𝐵(𝑥 − 1)
𝑥 − 1
 
 
2𝑥 − 1
𝑥 − 2
=
𝐴(𝑥 − 1)
𝑥 − 2
+ 𝐵 
(2 ⋅ 1) − 1
1 − 2
=
𝐴(1 − 1)
1 − 2
+ 𝐵 
𝐵 = −1 
2𝑥 − 1
𝑥2 − 3𝑥 + 2
=
3
𝑥 − 2
+
−1
𝑥 − 1
 
 
 
 
 
∫
2𝑥 − 1
𝑥2 − 3𝑥 + 2
𝑑𝑥 = ∫ (
3
𝑥 − 2
+
−1
𝑥 − 1
) 𝑑𝑥 
∫
3
𝑥 − 2
𝑑𝑥 − ∫
1
𝑥 − 1
𝑑𝑥 
 
∫
3
𝑥 − 2
𝑑𝑥 = 3 ⋅ ∫
1
𝑥 − 2
 𝑑𝑥 
𝑢 = 𝑥 − 2 𝑑𝑢 = 𝑑𝑥 
3 ⋅ ∫
1
𝑢
𝑑𝑢 = 3 ln |𝑢| 
3 ln|𝑥 − 2| = ln|(𝑥 − 2)3| 
 
∫
1
𝑥 − 1
𝑑𝑥 
𝑤 = 𝑥 − 1 𝑑𝑤 = 𝑑𝑥 
∫
1
𝑤
𝑑𝑤 = ln|𝑤| 
ln|𝑥 − 1| 
 
ln|(𝑥 − 2)3| − ln|𝑥 − 1| + 𝐶 
 
ln |
(𝑥 − 2)3
𝑥 − 1
| + 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
e) ∫ 3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 
𝑢 = 3𝑥 𝑑𝑢 = 3 𝑑𝑥 
𝑑𝑣 = 𝑒−2𝑥 𝑑𝑥 𝑣 = ∫ 𝑑𝑣 
 
∫ 𝑒−2𝑥𝑑𝑥 
𝑢 = −2𝑥 𝑑𝑢 = −2 𝑑𝑥 𝑑𝑥 =
−𝑑𝑢
2
 
∫ 𝑒𝑢 ⋅ (
−𝑑𝑢
2
) 
−
1
2
∫ 𝑒𝑢 𝑑𝑢 
−𝑒𝑢
2
 
𝑣 =
−𝑒−2𝑥
2
 
 
∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢 
 
∫ 3𝑥 ⋅ 𝑒−2𝑥𝑑𝑥 =
−3𝑥 ⋅ 𝑒−2𝑥
2
− ∫
−3 ⋅ 𝑒−2𝑥
2
𝑑𝑥 
∫ 3𝑥 ⋅ 𝑒−2𝑥𝑑𝑥 =
−3𝑥 ⋅ 𝑒−2𝑥
2
+
3
2
⋅ ∫ 𝑒−2𝑥𝑑𝑥 
 
∫ 𝑒−2𝑥𝑑𝑥 
𝑢 = −2𝑥 𝑑𝑢 = −2 𝑑𝑥 𝑑𝑥 =
−𝑑𝑢
2
 
∫ 𝑒𝑢 ⋅ (
−𝑑𝑢
2
) 
−
1
2
⋅ ∫ 𝑒𝑢 𝑑𝑢 
−𝑒𝑢
2
 
−𝑒−2𝑥
2
 
 
 
 
∫ 3𝑥 ⋅ 𝑒−2𝑥𝑑𝑥 =
−3𝑥 ⋅ 𝑒−2𝑥
2
+
3
2
⋅ (
−𝑒−2𝑥
2
) 
 
∫ 3𝑥 ⋅ 𝑒−2𝑥𝑑𝑥 =
−3𝑥 ⋅ 𝑒−2𝑥
2
+
−3𝑒−2𝑥
4
 
 
−
3𝑥 ⋅ 𝑒−2𝑥
2
−
3𝑒−2𝑥
4
+ 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
f) ∫
𝑥−1
1+𝑥2
⋅ cos(2𝜋) 𝑑𝑥 
 
cos(2𝜋) = 1 
∫
𝑥 − 1
1 + 𝑥2
 𝑑𝑥 
∫
𝑥
1 + 𝑥2
 𝑑𝑥 − ∫
1
1 + 𝑥2
 𝑑𝑥 
 
∫
𝑥
1 + 𝑥2
 𝑑𝑥 
 
𝑢 = 1 + 𝑥2 𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑥 =
𝑑𝑢
2𝑥
 
∫
𝑥
𝑢
𝑑𝑢
2𝑥
 
∫
𝑑𝑢
2𝑢
 
1
2
⋅ ∫
𝑑𝑢
𝑢
 
ln|𝑢|
2ln|1 + 𝑥2|
2
 
 
∫
1
1 + 𝑥2
 𝑑𝑥 = arctg(𝑥) 
 
∫
𝑥 − 1
1 + 𝑥2
⋅ cos(2𝜋) 𝑑𝑥 =
ln|1 + 𝑥2|
2
− arctg(𝑥) + 𝐶 
 
ln|1 + 𝑥2|
2
− arctg(𝑥) + 𝐶 
 
 
 
 
 
 
g) ∫ cos3(𝑥)𝑑𝑥 
∫ cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 
𝑢 = cos2(𝑥) 𝑑𝑢 = −2 cos(𝑥) ⋅ sen(𝑥) 𝑑𝑥 
𝑑𝑣 = cos(𝑥) 𝑑𝑥 𝑣 = sen(𝑥) 
 
∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢 
∫ cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) − ∫ −2 cos(𝑥) ⋅ sen(𝑥) ⋅ sen(𝑥) 𝑑𝑥 
∫ cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + ∫ 2 cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
∫ cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + 2 ∫ cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
∫ cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
 
𝑤 = sen(𝑥) 𝑑𝑤 = cos(𝑥) 𝑑𝑥 
𝑑𝑥 =
𝑑𝑤
cos(𝑥)
 
∫ cos(𝑥) ⋅ 𝑤2 
𝑑𝑤
cos(𝑥)
 
∫ 𝑤2 𝑑𝑤 
𝑤3
3
 
sen3(𝑥)
3
 
 
∫ cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + 2 ⋅
sen3(𝑥) 
3
+ 𝐶 
 
cos2(𝑥) ⋅ sen(𝑥) + 2 ⋅
sen3(𝑥) 
3
+ 𝐶 
cos2(𝑥) ⋅ sen(𝑥) +
2 sen3(𝑥) 
3
+ 𝐶 
 
 
 
3 cos2(𝑥) ⋅ sen(𝑥) + 2 sen3(𝑥)
3
+ 𝐶 
 2 sen3(𝑥) + 3 sen(𝑥) cos2(𝑥) 
3
+ C 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
h) ∫
𝑥
1+𝑥4
 𝑑𝑥 
 
∫
𝑥
1 + (𝑥2)2
 𝑑𝑥 
 
𝑢 = 𝑥2 𝑑𝑢 = 2𝑥 𝑑𝑥 
𝑑𝑥 =
𝑑𝑢
2𝑥
 
 
∫
𝑥
1 + 𝑢2
𝑑𝑢
2𝑥
 
 
∫
1
1 + 𝑢2
𝑑𝑢
2
 
 
1
2
∫
1
1 + 𝑢2
 𝑑𝑢 
 
1
2
⋅ arctg(𝑢) + 𝐶 
 
1
2
⋅ arctg(𝑥2) + 𝐶 
 
arctg(𝑥2)
2
+ 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
i) ∫ ln(𝑥 + 3) 𝑑𝑥 
𝑢 = 𝑥 + 3 𝑑𝑢 = 𝑑𝑥 
∫ ln(𝑢) 𝑑𝑢 
∫ ln(𝑢) = 𝑢 ⋅ ln(𝑢) − 𝑢 
(𝑥 + 3) ⋅ ln(𝑥 + 3) − (𝑥 + 3) 
(𝑥 + 3) ⋅ ln(𝑥 + 3) − 𝑥 − 3 + 𝐶 −3 + 𝐶 = 𝐶 
(𝑥 + 3) ⋅ ln(𝑥 + 3) − 𝑥 + 𝐶 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
j) ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 
 
𝑢 = 𝑒2𝑥 𝑑𝑢 = 2𝑒2𝑥 
𝑑𝑣 = cos(3𝑥) 𝑑𝑥 𝑣 = ∫ 𝑑𝑣 
∫ cos(3𝑥) 𝑑𝑥 
𝑤 = 3𝑥 𝑑𝑤 = 3 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
3
 
𝑣 = ∫ cos(𝑤)
𝑑𝑤
3
 
𝑣 =
sen(𝑤)
3
 
𝑣 =
sen(3𝑥)
3
 
 
 
∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
− ∫
sen(3𝑥) ⋅ 2𝑒2𝑥
3
 𝑑𝑥 
 
−
2
3
∫ sen(3𝑥) ⋅ 𝑒2𝑥𝑑𝑥 
𝑢 = 𝑒2𝑥 𝑑𝑢 = 2𝑒2𝑥 
𝑑𝑣 = sen(3𝑥) 𝑑𝑥 𝑣 = ∫ 𝑑𝑣 
 
 
 
 
 
 
 
 
 
 
 
 
𝑣 = ∫ sen(3𝑥) 𝑑𝑥 
ℎ = 3𝑥 𝑑ℎ = 3 𝑑𝑥 𝑑𝑥 =
𝑑ℎ
3
 
𝑣 = ∫ sen(ℎ) 
𝑑ℎ
3
 
𝑣 = −
cos(ℎ)
3
 
𝑣 = −
cos(3𝑥)
3
 
 
−
2
3
∫ sen(3𝑥) ⋅ 𝑒2𝑥𝑑𝑥 = −
2
3
(
−𝑒2𝑥 ⋅ cos(3𝑥)
3
) −
2
3
⋅ (− ∫
−2𝑒2𝑥 ⋅ cos(3𝑥)
3
𝑑𝑥) 
 
2𝑒2𝑥 ⋅ cos(3𝑥)
9
− ∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
𝑑𝑥 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
− ∫
sen(3𝑥) ⋅ 2𝑒2𝑥
3
 𝑑𝑥 
 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
− ∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
 𝑑𝑥 
 
∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
 𝑑𝑥 + ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
4
9
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 + ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
 
4
9
+ 1 =
13
9
 
 
 
 
 
𝑒2𝑥
13
⋅ [3 sen(3𝑥) + 2 cos(3𝑥)] + 𝐶 
13
9
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
9
13
⋅ (
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
) 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
9
13
⋅
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
9
13
⋅
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
3𝑒2𝑥 ⋅ sen(3𝑥)
13
+
2𝑒2𝑥 ⋅ cos(3𝑥)
13
 
 
∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥
13
⋅ [3 sen(3𝑥) + 2 cos(3𝑥)] + 𝐶 
 
 
 
 
 
k) ∫ 𝑒ln(𝑥
2) ⋅ 2𝑥 𝑑𝑥 
 
𝑒ln(𝑎) = 𝑎 
∫ 𝑥2 ⋅ 2𝑥 𝑑𝑥 
∫ 2𝑥3𝑑𝑥 =
2𝑥4
4
 
∫ 2𝑥3𝑑𝑥 =
𝑥4
2
 
 
𝑥4
4
+ 𝐶