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Evaluate the integral. (Use C for the constant of integration.)3x3 4x221x + 4 dx(x21)(x27)

Question

How do you integrate

3x3 + 4x2 + 21x + 4/
(x2 + 1)(x2 + 7)

by partial fractions

Evaluate the integral. (Use C for the constant of integration.)
3x3 4x221x + 4 dx
(x21)(x27)
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Evaluate the integral. (Use C for the constant of integration.) 3x3 4x221x + 4 dx (x21)(x27)

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Step 1

We need to evaluate the given integral using partial fractions.

3x34x221x +4
(x2 1)(x27)
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3x34x221x +4 (x2 1)(x27)

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Step 2

We can proceed by finding the partial fractions as

3x34221x + 4
сх + d
ах + b
x21
(x2 1)x27)
x2 7
1)(x2 +7)
Multiplying both sides by (x2
3x3
21x + 4
(ах + b)(x? + 7) + (сх + d)(х? + 1)
(ac)x3(b + d)x2 + (7a + c)x
(7b d)
and RHS we get,
By comparing the coefficients of LH
a c 3
bd 4
7ac 21
7bd 4
solving the above equation gives,
b 0
а 3 3,
c 0
d 4
3x34x
21x + 4
Зx
4
Thus,
x21x2 7
(x21)(x27)
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3x34221x + 4 сх + d ах + b x21 (x2 1)x27) x2 7 1)(x2 +7) Multiplying both sides by (x2 3x3 21x + 4 (ах + b)(x? + 7) + (сх + d)(х? + 1) (ac)x3(b + d)x2 + (7a + c)x (7b d) and RHS we get, By comparing the coefficients of LH a c 3 bd 4 7ac 21 7bd 4 solving the above equation gives, b 0 а 3 3, c 0 d 4 3x34x 21x + 4 Зx 4 Thus, x21x2 7 (x21)(x27)

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Step 3

Hence,

...
--
3x34x 21x 4
dx
(x2 1)(x27)
3x
dx
x21
4
dx
x27
3x
dx
x2 1
X
dx
x21
Let x2 1 = z
=> x dx = dz/2
dz
Inlzl =
3
dx
2
S
X
2Inzlnx?
Inlx21
x2 1
Z
4
dx
x2 7
4
x2 7
dx
Let x = 7u=> dx = V7du
1
dx
x2 7
4
du
4
tan u =
V7
X
-1
4
tan
=
_
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-- 3x34x 21x 4 dx (x2 1)(x27) 3x dx x21 4 dx x27 3x dx x2 1 X dx x21 Let x2 1 = z => x dx = dz/2 dz Inlzl = 3 dx 2 S X 2Inzlnx? Inlx21 x2 1 Z 4 dx x2 7 4 x2 7 dx Let x = 7u=> dx = V7du 1 dx x2 7 4 du 4 tan u = V7 X -1 4 tan = _

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