Question
Asked Oct 17, 2019
4) Solve the integral: Show work
10x2-48x + 160
dx
(x - 4)2 (x2 + 16)
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4) Solve the integral: Show work 10x2-48x + 160 dx (x - 4)2 (x2 + 16)

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Expert Answer

Step 1

Given,

10x2-48x160
dx
(x-4)2 (x2+16)
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10x2-48x160 dx (x-4)2 (x2+16)

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Step 2

Now, finding the partial fraction

10х2-48х+160
А
В
(x-4)2(x2+16)
(x-4)2
(x2+16)
А(x? + 16) + В(х — 4)2
10x2-48x + 160 =
AX216A Bx2 16B - 8xB
10x2
48x160
x2 (AB) 8xB + 16(A + B)
10x2-48x + 160 -
= X
Now, comparing the coefficient on both sides, we get
— 10 аnd — 8B
А+ В
-48
-48 » B = 6
So, -8B
. A +6
10 > A = 4
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10х2-48х+160 А В (x-4)2(x2+16) (x-4)2 (x2+16) А(x? + 16) + В(х — 4)2 10x2-48x + 160 = AX216A Bx2 16B - 8xB 10x2 48x160 x2 (AB) 8xB + 16(A + B) 10x2-48x + 160 - = X Now, comparing the coefficient on both sides, we get — 10 аnd — 8B А+ В -48 -48 » B = 6 So, -8B . A +6 10 > A = 4

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Step 3

Now,

...
10х2-48х+160
dx
(x-4)2 (x2+16)
4
6
dx
+
(x2+16)
(х-4)2
4
dx
(х-4)2
6
dx
(x2+16)
1
dx
(x2+16)
= 4 J(x- 4)-2dx + 6
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10х2-48х+160 dx (x-4)2 (x2+16) 4 6 dx + (x2+16) (х-4)2 4 dx (х-4)2 6 dx (x2+16) 1 dx (x2+16) = 4 J(x- 4)-2dx + 6

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Tagged in

Math

Calculus

Integration