∫
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 precedente 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
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)
