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∫
x
+
1
x
2
−
5
x
+
6
d
x
{\displaystyle \ \int _{}{\frac {x+1}{x^{2}-5x+6}}dx}
Si ha:
x
2
−
5
x
+
6
=
(
x
−
2
)
(
x
−
3
)
{\displaystyle \ x^{2}-5x+6=(x-2)(x-3)}
,
x
+
1
x
2
−
5
x
+
6
=
c
1
x
−
2
+
c
2
x
−
3
{\displaystyle {\frac {x+1}{x^{2}-5x+6}}={\frac {c_{1}}{x-2}}+{\frac {c_{2}}{x-3}}}
,
x
+
1
=
(
c
1
+
c
2
)
x
−
3
c
1
−
2
c
2
{\displaystyle \ x+1=(c_{1}+c_{2})x-3c_{1}-2c_{2}}
, da cui:
{
c
1
+
c
2
=
1
−
3
c
1
−
2
c
2
=
1
{\displaystyle \left\{{\begin{matrix}c_{1}+c_{2}=1\\-3c_{1}-2c_{2}=1\end{matrix}}\right.}
Risolvendo il sistema si ha:
c
1
=
−
3
{\displaystyle \ c_{1}=-3}
c
2
=
4
{\displaystyle \ c_{2}=4}
Quindi:
∫
x
+
1
x
2
−
5
x
+
6
d
x
=
∫
−
3
x
−
2
d
x
+
∫
4
x
−
3
d
x
=
l
o
g
(
x
−
3
)
4
(
x
−
2
)
3
{\displaystyle \int _{}{\frac {x+1}{x^{2}-5x+6}}dx=\ \int _{}{\frac {-3}{x-2}}dx+\ \int _{}{\frac {4}{x-3}}dx=log{\frac {(x-3)^{4}}{(x-2)^{3}}}}
∫
x
3
+
x
+
1
x
3
−
x
2
+
x
−
1
d
x
{\displaystyle \ \int _{}{\frac {x^{3}+x+1}{x^{3}-x^{2}+x-1}}dx}
Eseguendo la divisione si ha:
∫
x
3
+
x
+
1
x
3
−
x
2
+
x
−
1
=
1
+
x
2
+
2
x
3
−
x
2
+
x
−
1
=
1
+
x
2
+
2
(
x
−
1
)
(
x
2
+
1
)
{\displaystyle \ \int _{}{\frac {x^{3}+x+1}{x^{3}-x^{2}+x-1}}=1+{\frac {x^{2}+2}{x^{3}-x^{2}+x-1}}=1+{\frac {x^{2}+2}{(x-1)(x^{2}+1)}}}
Scomponendo la seconda frazione ottenuta e determinando le costanti come nell'esempio prescedente si trova:
x
2
+
2
x
3
−
x
2
+
x
−
1
=
c
1
x
−
1
+
c
2
x
+
c
2
x
2
+
1
=
3
2
1
x
−
1
−
1
2
x
+
1
x
2
+
1
{\displaystyle {\frac {x^{2}+2}{x^{3}-x^{2}+x-1}}={\frac {c_{1}}{x-1}}+{\frac {c_{2}x+c_{2}}{x^{2}+1}}={\frac {3}{2}}{\frac {1}{x-1}}-{\frac {1}{2}}{\frac {x+1}{x^{2}+1}}}
Quindi:
∫
x
3
+
x
+
1
x
3
−
x
2
+
x
−
1
d
x
=
x
+
3
2
l
o
g
(
x
−
1
)
−
1
4
l
o
g
(
x
2
+
1
)
−
1
2
a
r
c
t
a
n
g
(
x
)
=
{\displaystyle \ \int _{}{\frac {x^{3}+x+1}{x^{3}-x^{2}+x-1}}dx=x+{\frac {3}{2}}log(x-1)-{\frac {1}{4}}log(x^{2}+1)-{\frac {1}{2}}arc\ tang(x)=}
=
x
+
l
o
g
(
x
−
1
)
3
(
x
2
+
1
)
−
1
2
a
r
c
t
a
n
g
(
x
)
{\displaystyle =x+log{\sqrt {\frac {(x-1)^{3}}{{\sqrt {(}}x^{2}+1)}}}-{\frac {1}{2}}arc\ tang(x)}
∫
x
3
−
x
2
+
1
(
1
+
x
2
)
3
d
x
{\displaystyle \int {}{}{x^{3}-x^{2}+1 \over (1+x^{2})^{3}}dx}
Applicando la formula notevole
∫
A
(
x
)
(
a
x
2
+
b
)
n
d
x
=
∑
i
=
1
2
n
−
2
c
i
x
i
−
1
(
a
x
2
+
n
)
n
−
1
+
c
2
n
−
1
log
(
a
x
2
+
b
)
+
c
2
n
I
0
(
x
)
{\displaystyle \int {}{}{A(x) \over (ax^{2}+b)^{n}}dx={\sum _{i=1}^{2n-2}c_{i}x^{i-1} \over (ax^{2}+n)^{n-1}}+c_{2n-1}\log(ax^{2}+b)+c_{2n}I_{0}(x)}
Derivando i due membri, riducendo i risultati allo stesso denominatore e confrontando poi i numeratori, si trovano i valori:
c
1
=
1
4
c
2
=
−
1
2
c
3
=
3
4
c
4
=
−
1
4
c
5
=
0
c
6
=
1
4
{\displaystyle \ c_{1}={1 \over 4}\quad c_{2}={-1 \over 2}\quad c_{3}={3 \over 4}\quad c_{4}={-1 \over 4}\quad c_{5}=0\quad c_{6}={1 \over 4}}
∫
x
−
1
x
2
+
x
3
d
x
{\displaystyle \int {}{}{x-1 \over {\sqrt[{2}]{x}}+{\sqrt[{3}]{x}}}dx}
ponendo
:
x
=
t
6
d
x
=
6
t
5
d
t
s
i
h
a
{\displaystyle :\qquad x=t^{6}\qquad dx=6t^{5}dt\qquad si\ ha}
∫
x
−
1
x
2
+
x
3
d
x
=
6
∫
(
t
6
−
1
)
t
3
t
+
1
d
t
=
6
∫
(
t
5
−
t
4
+
t
3
−
t
2
+
t
−
1
)
t
3
d
t
{\displaystyle \int {}{}{x-1 \over {\sqrt[{2}]{x}}+{\sqrt[{3}]{x}}}dx=6\int {}{}{(t^{6}-1)t^{3} \over t+1}dt=6\int {}{}(t^{5}-t^{4}+t^{3}-t^{2}+t-1)t^{3}dt}
=
6
∫
(
t
8
−
t
7
+
t
6
−
t
5
+
t
4
−
t
3
)
d
t
=
2
3
t
9
−
3
4
t
8
+
6
7
t
7
−
t
6
+
6
5
t
5
−
3
2
t
4
=
{\displaystyle =6\int {}{}(t^{8}-t^{7}+t^{6}-t^{5}+t^{4}-t^{3})dt={2 \over 3}t^{9}-{3 \over 4}t^{8}+{6 \over 7}t^{7}-t^{6}+{6 \over 5}t^{5}-{3 \over 2}t^{4}=}
=
2
3
x
3
2
−
3
4
x
4
3
+
6
7
x
7
6
−
x
+
6
5
x
5
6
−
3
2
x
2
3
.
{\displaystyle ={2 \over 3}x^{3 \over 2}-{3 \over 4}x^{4 \over 3}+{6 \over 7}x^{7 \over 6}-x+{6 \over 5}x^{5 \over 6}-{3 \over 2}x^{2 \over 3}.}
∫
a
4
x
2
+
x
−
3
2
d
x
{\displaystyle \int _{}{}{a \over {\sqrt[{2}]{4x^{2}+x-3}}}dx}
Si può eseguire con la posizione:
4
x
2
+
x
−
3
2
=
t
−
2
x
{\displaystyle {\sqrt[{2}]{4x^{2}+x-3}}=t-2x}
t
∫
x
4
x
2
+
x
−
3
2
d
x
=
c
1
4
x
2
+
x
−
3
2
+
c
2
∫
d
x
4
x
2
+
x
−
3
2
.
{\displaystyle \int _{}{}{x \over {\sqrt[{2}]{4x^{2}+x-3}}}dx=c_{1}{\sqrt[{2}]{4x^{2}+x-3}}+c_{2}\int {}{}{dx \over {\sqrt[{2}]{4x^{2}+x-3}}}.}
Derivando i due membri si ha:
x
4
x
2
+
x
−
3
2
=
c
1
(
4
x
+
1
2
)
4
x
2
+
x
−
3
2
+
c
2
4
x
2
+
x
−
3
2
,
{\displaystyle {x \over {\sqrt[{2}]{4x^{2}+x-3}}}={c_{1}(4x+{1 \over 2}) \over {\sqrt[{2}]{4x^{2}+x-3}}}+{c_{2} \over {\sqrt[{2}]{4x^{2}+x-3}}},}
da cui risulta
:
c
1
=
1
4
,
c
2
=
−
1
8
.
{\displaystyle :\qquad c_{1}={1 \over 4},\qquad c_{2}=-{1 \over 8}.}
∫
x
2
−
3
x
+
1
3
x
+
2
3
d
x
{\displaystyle \int _{}{}{x^{2}-3x+1 \over {\sqrt[{3}]{3x+2}}}dx}
Applicando la formula notevole
D
)
1
{\displaystyle \ D)\ 1}
∫
x
2
x
6
+
1
2
d
x
{\displaystyle \int {}{}{\sqrt[{2}]{x}}{\sqrt[{2}]{{\sqrt[{6}]{x}}+1}}\ dx}
essendo
m
=
1
2
n
=
1
6
,
{\displaystyle \ m={1 \over 2}\quad n={1 \over 6,}}
m
+
1
n
=
(
1
2
+
1
)
:
1
6
=
9
,
{\displaystyle \ {m+1 \over n}=({1 \over 2}+1)\ :{1 \over 6}=9\ ,}
e perciò ponendo :
x
6
+
1
=
t
{\displaystyle {\sqrt[{6}]{x}}+1=t}
x
=
(
t
−
1
)
6
,
d
x
=
6
(
t
−
1
)
5
d
t
,
{\displaystyle \ x=(t-1)^{6}\ ,\quad dx=6(t-1)^{5}dt\ ,}
∫
x
2
x
6
+
1
2
d
x
=
6
∫
(
t
−
1
)
8
t
2
d
t
,
{\displaystyle \int {\sqrt[{2}]{x}}{\sqrt[{2}]{{\sqrt[{6}]{x}}+1}}\ dx=6\int {(t-1)^{8}}{\sqrt[{2}]{t}}\ dt\ ,}
∫
sin
3
x
d
x
{\displaystyle \int {}{}\sin ^{3}{x}dx}
∫
sin
3
x
d
x
=
−
∫
sin
2
x
d
cos
x
=
−
∫
(
1
−
cos
2
x
)
d
cos
x
=
−
(
cos
x
−
cos
3
x
3
)
.
{\displaystyle \int \sin ^{3}xdx=-\int \sin ^{2}x\ d\cos x=-\int (1-\cos ^{2}x)d\cos x=-(\cos x-{\cos ^{3}x \over 3})\ .}
∫
cos
3
x
d
x
=
∫
(
1
−
sin
2
x
)
d
sin
x
=
sin
x
−
sin
3
x
3
.
{\displaystyle \int \cos ^{3}x\ dx=\int (1-\sin ^{2}x)\ d\sin x=\sin x-{\sin ^{3}x \over 3}\ .}
![{\displaystyle \int {}{}{x-1 \over {\sqrt[{2}]{x}}+{\sqrt[{3}]{x}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac4c7a110b0b1481fd69418db4598dd3370e3ce6)
![{\displaystyle \int {}{}{x-1 \over {\sqrt[{2}]{x}}+{\sqrt[{3}]{x}}}dx=6\int {}{}{(t^{6}-1)t^{3} \over t+1}dt=6\int {}{}(t^{5}-t^{4}+t^{3}-t^{2}+t-1)t^{3}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a044c52394adb02219136d030b499f7adf94d10)
![{\displaystyle \int _{}{}{a \over {\sqrt[{2}]{4x^{2}+x-3}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50f70b9b1c468f2eff05d0f29648918c412186b0)
![{\displaystyle {\sqrt[{2}]{4x^{2}+x-3}}=t-2x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1319f3f63600b3352b46567e3c0934467633c86f)
![{\displaystyle \int _{}{}{x \over {\sqrt[{2}]{4x^{2}+x-3}}}dx=c_{1}{\sqrt[{2}]{4x^{2}+x-3}}+c_{2}\int {}{}{dx \over {\sqrt[{2}]{4x^{2}+x-3}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa1ea7f792607c946a0e45ddc14e994183786d9b)
![{\displaystyle {x \over {\sqrt[{2}]{4x^{2}+x-3}}}={c_{1}(4x+{1 \over 2}) \over {\sqrt[{2}]{4x^{2}+x-3}}}+{c_{2} \over {\sqrt[{2}]{4x^{2}+x-3}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baf3b10fda394713406626d514a2fdfa0a4d9fad)
![{\displaystyle \int _{}{}{x^{2}-3x+1 \over {\sqrt[{3}]{3x+2}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a874e7e3fb928d0b54754add8f13d06babd3717)
![{\displaystyle \int {}{}{\sqrt[{2}]{x}}{\sqrt[{2}]{{\sqrt[{6}]{x}}+1}}\ dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a922d669c0f9ccc42210eedabf0efeeaf5f71d0c)
![{\displaystyle {\sqrt[{6}]{x}}+1=t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/080dc4d5603a058b051176db4b64741dcf327a14)
![{\displaystyle \int {\sqrt[{2}]{x}}{\sqrt[{2}]{{\sqrt[{6}]{x}}+1}}\ dx=6\int {(t-1)^{8}}{\sqrt[{2}]{t}}\ dt\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7239c92c7f8b3b73923756debbe372f074992f8a)
