(3+7x) √(9 - 3x ) dx
Answers
Answer:
Solution
verified
Verified by Toppr
Let I=∫
7x+9
3x+5
dx
Let 3x+5=λ(7x+9)+μ
On equating the coefficeints of like powers of x on both sides, we get
⇒3=7λ⇒λ=
7
3
⇒5=9λ+μ⇒μ=5−9λ=5−9×
7
3
=
7
35−27
=
7
8
Replacing 3x+5 by λ(7x+9)+μ in the given integral, we get
I=∫
7x+9
λ(7x+9)+μ
dx
=∫
7x+9
7
3
(7x+9)+
7
8
dx
=
7
3
∫
7x+9
(7x+9)
dx+
7
8
∫
7x+9
1
dx
=
7
3
∫(7x+9)
1−
2
1
dx+
7
8
∫(7x+9)
−
2
1
dx
=
7
3
∫(7x+9)
2
1
dx+
7
8
∫(7x+9)
−
2
1
dx
We know that ∫(ax+b)
n
=
a(n+1)
1
(ax+b)
n+1
=
7
3
×
7
1
2
1
+1
(7x+9)
2
1
+1
+
7
8
×
7
1
−
2
1
+1
(7x+9)
−
2
1
+1
+c
=
49
3
2
3
(7x+9)
2
3
+
49
8
2
1
(7x+9)
2
1
+c
=
49
2
(7x+9)
2
3
+
49
16
(7x+9)
2
1
+c
=
49
2
(7x+9)(7x+9)
2
1
+
49
16
(7x+9)
2
1
+c
=
49
2
[(7x+9)+8](7x+9)
2
1
+c
=
49
2
(7x+17)(7x+9)
2
1
+c
Step-by-step explanation:
pls make me Brainlinest