tabel integral .
∫
(
a
x
+
b
)
n
d
x
{\displaystyle \int (ax+b)^{n}dx}
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
(untuk
n
≠
−
1
)
{\displaystyle ={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{(untuk }}n\neq -1{\mbox{)}}\,\!}
∫
1
a
x
+
b
d
x
{\displaystyle \int {\frac {1}{ax+b}}dx}
=
1
a
ln
|
a
x
+
b
|
{\displaystyle ={\frac {1}{a}}\ln \left|ax+b\right|}
∫
x
(
a
x
+
b
)
n
d
x
{\displaystyle \int x(ax+b)^{n}dx}
=
a
(
n
+
1
)
x
−
b
a
2
(
n
+
1
)
(
n
+
2
)
(
a
x
+
b
)
n
+
1
(untuk
n
∉
{
−
1
,
−
2
}
)
{\displaystyle ={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{(untuk }}n\not \in \{-1,-2\}{\mbox{)}}}
∫
x
a
x
+
b
d
x
{\displaystyle \int {\frac {x}{ax+b}}dx}
=
x
a
−
b
a
2
ln
|
a
x
+
b
|
{\displaystyle ={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|}
∫
x
(
a
x
+
b
)
2
d
x
{\displaystyle \int {\frac {x}{(ax+b)^{2}}}dx}
=
b
a
2
(
a
x
+
b
)
+
1
a
2
ln
|
a
x
+
b
|
{\displaystyle ={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|}
∫
x
(
a
x
+
b
)
n
d
x
{\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx}
=
a
(
1
−
n
)
x
−
b
a
2
(
n
−
1
)
(
n
−
2
)
(
a
x
+
b
)
n
−
1
(untuk
n
∉
{
1
,
2
}
)
{\displaystyle ={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}\qquad {\mbox{(untuk }}n\not \in \{1,2\}{\mbox{)}}}
∫
x
2
a
x
+
b
d
x
{\displaystyle \int {\frac {x^{2}}{ax+b}}dx}
=
1
a
3
(
(
a
x
+
b
)
2
2
−
2
b
(
a
x
+
b
)
+
b
2
ln
|
a
x
+
b
|
)
{\displaystyle ={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}
∫
x
2
(
a
x
+
b
)
2
d
x
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx}
=
1
a
3
(
a
x
+
b
−
2
b
ln
|
a
x
+
b
|
−
b
2
a
x
+
b
)
{\displaystyle ={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}
∫
x
2
(
a
x
+
b
)
3
d
x
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx}
=
1
a
3
(
ln
|
a
x
+
b
|
+
2
b
a
x
+
b
−
b
2
2
(
a
x
+
b
)
2
)
{\displaystyle ={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)}
∫
x
2
(
a
x
+
b
)
n
d
x
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx}
=
1
a
3
(
−
(
a
x
+
b
)
3
−
n
(
n
−
3
)
+
2
b
(
a
+
b
)
2
−
n
(
n
−
2
)
−
b
2
(
a
x
+
b
)
1
−
n
(
n
−
1
)
)
(untuk
n
∉
{
1
,
2
,
3
}
)
{\displaystyle ={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(a+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)\qquad {\mbox{(untuk }}n\not \in \{1,2,3\}{\mbox{)}}}
∫
1
x
(
a
x
+
b
)
d
x
{\displaystyle \int {\frac {1}{x(ax+b)}}dx}
=
−
1
b
ln
|
a
x
+
b
x
|
{\displaystyle =-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}
∫
1
x
2
(
a
x
+
b
)
d
x
{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}dx}
=
−
1
b
x
+
a
b
2
ln
|
a
x
+
b
x
|
{\displaystyle =-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}
∫
1
x
2
(
a
x
+
b
)
2
d
x
{\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}dx}
=
−
a
(
1
b
2
(
a
x
+
b
)
+
1
a
b
2
x
−
2
b
3
ln
|
a
x
+
b
x
|
)
{\displaystyle =-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)}
∫
1
x
2
+
a
2
d
x
{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}dx}
=
1
a
arctan
x
a
{\displaystyle ={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!}
∫
1
x
2
−
a
2
d
x
=
{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}dx=}
−
1
a
a
r
c
t
a
n
h
x
a
=
1
2
a
ln
a
−
x
a
+
x
(untuk
|
x
|
<
|
a
|
)
{\displaystyle -{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}\qquad {\mbox{(untuk }}|x|<|a|{\mbox{)}}\,\!}
−
1
a
a
r
c
c
o
t
h
x
a
=
1
2
a
ln
x
−
a
x
+
a
(untuk
|
x
|
>
|
a
|
)
{\displaystyle -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{(untuk }}|x|>|a|{\mbox{)}}\,\!}
∫
1
a
x
2
+
b
x
+
c
d
x
=
{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx=}
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(untuk
4
a
c
−
b
2
>
0
)
{\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}4ac-b^{2}>0{\mbox{)}}}
−
2
b
2
−
4
a
c
a
r
c
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
=
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
(untuk
4
a
c
−
b
2
<
0
)
{\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|\qquad {\mbox{(untuk }}4ac-b^{2}<0{\mbox{)}}}
−
2
2
a
x
+
b
(untuk
4
a
c
−
b
2
=
0
)
{\displaystyle -{\frac {2}{2ax+b}}\qquad {\mbox{(untuk }}4ac-b^{2}=0{\mbox{)}}}
∫
x
a
x
2
+
b
x
+
c
d
x
{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx}
=
1
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
b
2
a
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle ={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}}
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx=}
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
2
a
n
−
b
m
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(untuk
4
a
c
−
b
2
>
0
)
{\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}4ac-b^{2}>0{\mbox{)}}}
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
2
a
n
−
b
m
a
b
2
−
4
a
c
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
(untuk
4
a
c
−
b
2
<
0
)
{\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(untuk }}4ac-b^{2}<0{\mbox{)}}}
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
(
2
a
x
+
b
)
(untuk
4
a
c
−
b
2
=
0
)
{\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}\,\,\,\,\,\,\,\,\,\,\qquad {\mbox{(untuk }}4ac-b^{2}=0{\mbox{)}}}
∫
1
(
a
x
2
+
b
x
+
c
)
n
d
x
=
2
a
x
+
b
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
(
2
n
−
3
)
2
a
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
{\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!}
∫
x
(
a
x
2
+
b
x
+
c
)
n
d
x
=
b
x
+
2
c
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
b
(
2
n
−
3
)
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx\,\!}
∫
1
x
(
a
x
2
+
b
x
+
c
)
d
x
=
1
2
c
ln
|
x
2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
1
a
x
2
+
b
x
+
c
d
x
{\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}dx}
Fungsi rasional apapun dapat diintegrasikan melalui persamaan-persamaan di atas dengan memanfaatkan integrasi parsial , dengan menguraikan fungsi rasional menjadi penjumlahan fungsi-fungsi dalam bentuk
e
x
+
f
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle {\frac {ex+f}{\left(ax^{2}+bx+c\right)^{n}}}}
Bacaan lebih lanjut
Kurnianingsih, Sri (2007). Matematika SMA dan MA 3A Untuk Kelas XII Semester 1 Program IPA . Jakarta: Esis/Erlangga. ISBN 979-734-504-1 . (Indonesia) Kurnianingsih, Sri (2007). Matematika SMA dan MA 3A Untuk Kelas XII Semester 1 Program IPS . Jakarta: Esis/Erlangga. ISBN 979-734-567-X . (Indonesia)
