Integrate Sqrt[x^2 + 4 x + 8]/(x + 2)
In[7]:=
Indefinite integrals:
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8 + 4 x + x2
á
2+x
Hx + 2L2 + 4 + 2 logHx + 2L - 2 log
âx‡
Hx + 2L2 + 4 + 2 + constant
Possible intermediate steps:
Take the integral:
8 + 4 x + x2
á
2+x
âx
x2 + 4 x + 8
For the integrand
, complete the square:
x+2
Hx + 2L2 + 4
‡á
âx
x+2
Hx + 2L2 + 4
For the integrand
x+2
u2 + 4
‡á
u
âu
u2 + 4
For the integrand
u
, substitute u ‡ 2 tanHsL and
âu ‡ 2 sec2 HsL âs. Then
‡4à
, substitute u ‡ x + 2 and âu ‡ â x:
1
2
u2 + 4 =
u
4 tan2 HsL + 4 = 2 secHsL and s ‡ tan-1 K O:
2
cscHsL sec2 HsL â s
Factor out constants:
‡ 2 à cscHsL sec2 HsL â s
For the integrand cscHsL sec2 HsL, use the trigonometric identity sec2 HsL ‡ tan2 HsL + 1:
‡ 2 à Itan2 HsL + 1M cscHsL â s
2
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Expanding the integrand Itan2 HsL + 1M cscHsL gives cscHsL + tanHsL secHsL:
‡ 2 à HcscHsL + tanHsL secHsLL â s
Integrate the sum term by term:
‡ 2 à cscHsL â s + 2 à tanHsL secHsL â s
For the integrand tanHsL secHsL, substitute p ‡ secHsL and â p ‡ tanHsL secHsLâs:
‡ 2 à 1 â p + 2 à cscHsL â s
The integral of 1 is p:
‡ 2 p + 2 à cscHsL â s
The integral of cscHsL is -logHcotHsL + cscHsLL:
‡ 2 p - 2 logHcotHsL + cscHsLL + constant
Substitute back for p ‡ secHsL:
‡ 2 secHsL - 2 logHcotHsL + cscHsLL + constant
u
Substitute back for s ‡ tan-1 K O:
2
‡
u2 + 4 - 2 log
u2 + 4 + 2
+ constant
u
Substitute back for u ‡ x + 2:
‡
2
Hx + 2L2 + 4 + 2
Hx + 2L + 4 - 2 log
+ constant
x+2
Factor the answer a different way:
Answer:
‡
x2 + 4 x + 8 - 2 log
Hx + 2L2 + 4 + 2
+ constant
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x2 + 4 x + 8 - 2 log
‡
+ constant
x+2
logHxL is the natural logarithm »
secHxL is the secant function »
-1
tan HxL is the inverse tangent function »
cscHxL is the cosecant function »
cotHxL is the cotangent function »
Plots of the integral:
Complex-valued plot È
Æ
Complex-valued plot È
Æ
5
x
-3
-2
1
-1
-5
real part
imaginary part
-10
min
max
10
5
x
-10
5
-5
10
real part
imaginary part
-5
min
max
Alternate forms of the integral:
x2 + 4 x + 8 - 2 log
x2 + 4 x + 8 + 2 + 2 logHx + 2L + constant
x2 + 4 x + 8 + 2 logHx + 2L - log
Hx + 2L2 + 4 + 2 + constant
x2 + 4 x + 8 + 2 logHx + 2L - log
x2 + 4 x + 8 + 2 + constant
Series expansion of the integral at x=-2:
H2 logHx + 2L + 2 - 2 logH4LL +
1
8
Hx + 2L2 + OIHx + 2L4 M
Series expansion of the integral at x=-2-2 i:
2
-ä Π +
4
-1 Hx + H2 + 2 äLL32 -
3
H-1L34 Hx + H2 + 2 äLL52 + OIHx + H2 + 2 äLL72 M
3
4
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4
-ä Π +
3
-1 Hx + H2 + 2 äLL32 -
20
H-1L34 Hx + H2 + 2 äLL52 + OIHx + H2 + 2 äLL72 M
Series expansion of the integral at x=-2+2 i:
2
äΠ3
H-1L34 Hx + H2 - 2 äLL32 +
3
4
20
-1 Hx + H2 - 2 äLL52 + OIHx + H2 - 2 äLL72 M
Series expansion of the integral at x=0:
2 - 2 logI1 +
I2
2 MM +
x2
2 x-
5 x3
+
4
2
48
+ OIx4 M
2
Series expansion of the integral at x=¥:
x+2-
2
4
+
x
x2
+O
1
3
x
Definite integral after subtraction of diverging parts:
More digits
¥
à
0
8 + 4 x + x2
2+x
-1 âx‡2-2
2 + 2 sinh-1 H1L » 0.93432
sinh-1 HxL is the inverse hyperbolic sine function »