Integration by parts (IBP) is a method of integration with the formula:
:
∫
a
b
u
(
x
)
v
′
(
x
)
d
x
=
[
u
(
x
)
v
(
x
)
]
a
b
−
∫
a
b
u
′
(
x
)
v
(
x
)
d
x
=
u
(
b
)
v
(
b
)
−
u
(
a
)
v
(
a
)
−
∫
a
b
u
′
(
x
)
v
(
x
)
d
x
.
{\displaystyle {\begin{aligned}\int _{a}^{b}u(x)v'(x)dx&={\Big [}u(x)v(x){\Big ]}_{a}^{b}-\int _{a}^{b}u'(x)v(x)dx\\[6pt]&=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)dx.\end{aligned}}}
Or more compactly,
∫
a
b
u
d
v
=
u
v
|
a
b
−
∫
a
b
v
d
u
{\displaystyle \int _{a}^{b}udv=uv|_{a}^{b}-\int _{a}^{b}vdu}
or without bounds
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int udv=uv-\int vdu}
where
u
{\displaystyle u}
and
v
{\displaystyle v}
are functions of a variable, for instance,
x
{\displaystyle x}
, giving
u
(
x
)
{\displaystyle u(x)}
and
v
(
x
)
{\displaystyle v(x)}
.
d
d
x
u
(
x
)
→
d
u
d
x
=
u
′
(
x
)
→
u
′
(
x
)
d
x
=
d
u
{\displaystyle {\frac {d}{dx}}u(x)\rightarrow {\frac {du}{dx}}=u'(x)\rightarrow u'(x)dx=du}
and
∫
d
v
=
∫
v
(
x
)
=
v
{\displaystyle \int dv=\int v(x)=v}
Note:
d
v
{\displaystyle dv}
is whatever terms are
not included as
u
{\displaystyle u}
.
A rule of thumb has been proposed, consisting of choosing
u
{\displaystyle u}
as the function that comes first in the following list:
I – inverse trigonometric functions:
arctan
(
x
)
,
arcsec
(
x
)
,
{\displaystyle \arctan(x),\ \operatorname {arcsec}(x),}
etc.
L – logarithmic functions:
ln
(
x
)
,
log
b
(
x
)
,
{\displaystyle \ln(x),\ \log _{b}(x),}
etc.
A – polynomials:
x
2
,
3
x
50
,
{\displaystyle x^{2},\ 3x^{50},}
etc.
T – trigonometric functions:
sin
(
x
)
,
tan
(
x
)
,
{\displaystyle \sin(x),\ \tan(x),}
etc.
E – exponential functions:
e
x
,
19
x
,
{\displaystyle e^{x},\ 19^{x},}
etc.
The theorem can be derived as follows. For two continuously differentiable functions
u
(
x
)
{\displaystyle u(x)}
and
v
(
x
)
{\displaystyle v(x)}
, the product rule states:
(
u
(
x
)
v
(
x
)
)
′
=
v
(
x
)
u
′
(
x
)
+
u
(
x
)
v
′
(
x
)
.
{\displaystyle {\Big (}u(x)v(x){\Big )}'\ =\ v(x)u'(x)+u(x)v'(x).}
Integrating both sides with respect to
x
{\displaystyle x}
,
∫
(
u
(
x
)
v
(
x
)
)
′
d
x
=
∫
u
′
(
x
)
v
(
x
)
d
x
+
∫
u
(
x
)
v
′
(
x
)
d
x
,
{\displaystyle \int {\Big (}u(x)v(x){\Big )}'\,dx\ =\ \int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx,}
and noting that an indefinite integral is an antiderivative gives
u
(
x
)
v
(
x
)
=
∫
u
′
(
x
)
v
(
x
)
d
x
+
∫
u
(
x
)
v
′
(
x
)
d
x
,
{\displaystyle u(x)v(x)\ =\ \int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx,}
where we neglect writing the constant of integration. This yields the formula for integration by parts:
∫
u
(
x
)
v
′
(
x
)
d
x
=
u
(
x
)
v
(
x
)
−
∫
u
′
(
x
)
v
(
x
)
d
x
,
{\displaystyle \int u(x)v'(x)\,dx\ =\ u(x)v(x)-\int u'(x)v(x)\,dx,}
or in terms of the differentials
d
u
=
u
′
(
x
)
d
x
,
d
v
=
v
′
(
x
)
d
x
,
{\displaystyle \ du=u'(x)\,dx,\ \ dv=v'(x)\,dx,\quad }
∫
u
(
x
)
d
v
=
u
(
x
)
v
(
x
)
−
∫
v
(
x
)
d
u
.
{\displaystyle \int u(x)\,dv\ =\ u(x)v(x)-\int v(x)\,du.}
This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two values x = a and x = b and applying the fundamental theorem of calculus gives the definite integral version:
∫
a
b
u
(
x
)
v
′
(
x
)
d
x
=
u
(
b
)
v
(
b
)
−
u
(
a
)
v
(
a
)
−
∫
a
b
u
′
(
x
)
v
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}u(x)v'(x)\,dx\ =\ u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)\,dx.}
Functions multiplied by one and itself
edit
Given
I
=
∫
ln
(
x
)
⋅
1
d
x
{\displaystyle I=\int \ln(x)\cdot 1\ dx\ }
edit
The first example is ∫ ln(x ) dx . We write this as:
I
=
∫
ln
(
x
)
⋅
1
d
x
.
{\displaystyle I=\int \ln(x)\cdot 1\ dx\ .}
Let:
u
=
ln
(
x
)
⇒
d
u
=
d
x
x
{\displaystyle u=\ln(x)\ \Rightarrow \ du={\frac {dx}{x}}}
d
v
=
d
x
⇒
v
=
x
{\displaystyle dv=dx\ \Rightarrow \ v=x}
then:
∫
ln
(
x
)
d
x
=
x
ln
(
x
)
−
∫
x
x
d
x
=
x
ln
(
x
)
−
∫
1
d
x
=
x
ln
(
x
)
−
x
+
C
{\displaystyle {\begin{aligned}\int \ln(x)\ dx&=x\ln(x)-\int {\frac {x}{x}}\ dx\\&=x\ln(x)-\int 1\ dx\\&=x\ln(x)-x+C\end{aligned}}}
where C is the constant of integration.
Given
I
=
∫
arctan
(
x
)
d
x
.
{\displaystyle I=\int \arctan(x)\ dx.}
edit
The second example is the inverse tangent function arctan(x ):
I
=
∫
arctan
(
x
)
d
x
.
{\displaystyle I=\int \arctan(x)\ dx.}
Rewrite this as
∫
arctan
(
x
)
⋅
1
d
x
.
{\displaystyle \int \arctan(x)\cdot 1\ dx.}
Now let:
u
=
arctan
(
x
)
⇒
d
u
=
d
x
1
+
x
2
{\displaystyle u=\arctan(x)\ \Rightarrow \ du={\frac {dx}{1+x^{2}}}}
d
v
=
d
x
⇒
v
=
x
{\displaystyle dv=dx\ \Rightarrow \ v=x}
then
∫
arctan
(
x
)
d
x
=
x
arctan
(
x
)
−
∫
x
1
+
x
2
d
x
=
x
arctan
(
x
)
−
ln
(
1
+
x
2
)
2
+
C
{\displaystyle {\begin{aligned}\int \arctan(x)\ dx&=x\arctan(x)-\int {\frac {x}{1+x^{2}}}\ dx\\[8pt]&=x\arctan(x)-{\frac {\ln(1+x^{2})}{2}}+C\end{aligned}}}
using a combination of the inverse chain rule method and the natural logarithm integral condition.
Polynomials and trigonometric functions
edit
In order to calculate
I
=
∫
x
cos
(
x
)
d
x
,
{\displaystyle I=\int x\cos(x)\ dx\ ,}
let:
u
=
x
⇒
d
u
=
d
x
{\displaystyle u=x\ \Rightarrow \ du=dx}
d
v
=
cos
(
x
)
d
x
⇒
v
=
∫
cos
(
x
)
d
x
=
sin
(
x
)
{\displaystyle dv=\cos(x)\ dx\ \Rightarrow \ v=\int \cos(x)\ dx=\sin(x)}
then:
∫
x
cos
(
x
)
d
x
=
∫
u
d
v
=
u
⋅
v
−
∫
v
d
u
=
x
sin
(
x
)
−
∫
sin
(
x
)
d
x
=
x
sin
(
x
)
+
cos
(
x
)
+
C
,
{\displaystyle {\begin{aligned}\int x\cos(x)\ dx&=\int u\ dv\\&=u\cdot v-\int v\,du\\&=x\sin(x)-\int \sin(x)\ dx\\&=x\sin(x)+\cos(x)+C,\end{aligned}}}
where C is a constant of integration.
For higher powers of x in the form
∫
x
n
e
x
d
x
,
∫
x
n
sin
(
x
)
d
x
,
∫
x
n
cos
(
x
)
d
x
,
{\displaystyle \int x^{n}e^{x}\ dx,\ \int x^{n}\sin(x)\ dx,\ \int x^{n}\cos(x)\ dx\ ,}
repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of x by one.
∫
x
3
e
x
2
d
x
,
{\displaystyle \int x^{3}e^{x^{2}}\,dx,}
one would set
u
=
x
2
,
d
v
=
x
⋅
e
x
2
d
x
,
{\displaystyle u=x^{2},\quad dv=x\cdot e^{x^{2}}\,dx,}
so that
d
u
=
2
x
d
x
,
v
=
e
x
2
2
.
{\displaystyle du=2x\,dx,\quad v={\frac {e^{x^{2}}}{2}}.}
Then
∫
x
3
e
x
2
d
x
=
∫
(
x
2
)
(
x
e
x
2
)
d
x
=
∫
u
d
v
=
u
v
−
∫
v
d
u
=
x
2
e
x
2
2
−
∫
x
e
x
2
d
x
.
{\displaystyle \int x^{3}e^{x^{2}}\,dx=\int (x^{2})\left(xe^{x^{2}}\right)\,dx=\int u\,dv=uv-\int v\,du={\frac {x^{2}e^{x^{2}}}{2}}-\int xe^{x^{2}}\,dx.}
Finally, this results in
∫
x
3
e
x
2
d
x
=
e
x
2
(
x
2
−
1
)
2
+
C
.
{\displaystyle \int x^{3}e^{x^{2}}\,dx={\frac {e^{x^{2}}(x^{2}-1)}{2}}+C.}
I
=
∫
e
x
cos
(
x
)
d
x
.
{\displaystyle I=\int e^{x}\cos(x)\ dx.}
Here, integration by parts is performed twice. First let
u
=
cos
(
x
)
⇒
d
u
=
−
sin
(
x
)
d
x
{\displaystyle u=\cos(x)\ \Rightarrow \ du=-\sin(x)\ dx}
d
v
=
e
x
d
x
⇒
v
=
∫
e
x
d
x
=
e
x
{\displaystyle dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}}
then:
∫
e
x
cos
(
x
)
d
x
=
e
x
cos
(
x
)
+
∫
e
x
sin
(
x
)
d
x
.
{\displaystyle \int e^{x}\cos(x)\ dx=e^{x}\cos(x)+\int e^{x}\sin(x)\ dx.}
Now, to evaluate the remaining integral, we use integration by parts again, with:
u
=
sin
(
x
)
⇒
d
u
=
cos
(
x
)
d
x
{\displaystyle u=\sin(x)\ \Rightarrow \ du=\cos(x)\ dx}
d
v
=
e
x
d
x
⇒
v
=
∫
e
x
d
x
=
e
x
.
{\displaystyle dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}.}
Then:
∫
e
x
sin
(
x
)
d
x
=
e
x
sin
(
x
)
−
∫
e
x
cos
(
x
)
d
x
.
{\displaystyle \int e^{x}\sin(x)\ dx=e^{x}\sin(x)-\int e^{x}\cos(x)\ dx.}
Putting these together,
∫
e
x
cos
(
x
)
d
x
=
e
x
cos
(
x
)
+
e
x
sin
(
x
)
−
∫
e
x
cos
(
x
)
d
x
.
{\displaystyle \int e^{x}\cos(x)\ dx=e^{x}\cos(x)+e^{x}\sin(x)-\int e^{x}\cos(x)\ dx.}
The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get
2
∫
e
x
cos
(
x
)
d
x
=
e
x
[
sin
(
x
)
+
cos
(
x
)
]
+
C
,
{\displaystyle 2\int e^{x}\cos(x)\ dx=e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C,}
which rearranges to
∫
e
x
cos
(
x
)
d
x
=
1
2
e
x
[
sin
(
x
)
+
cos
(
x
)
]
+
C
′
{\displaystyle \int e^{x}\cos(x)\ dx={\frac {1}{2}}e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C'}
1)
∫
2
x
cos
(
5
−
13
x
)
d
x
{\displaystyle \int 2x\cos(5-13x)dx}
2)
∫
x
cos
(
x
)
d
x
{\displaystyle \int x\cos(x)dx}
3)
∫
x
3
sin
(
3
x
)
d
x
{\displaystyle \int x^{3}\sin(3x)dx}
4)
∫
e
2
x
cos
(
3
x
)
d
x
{\displaystyle \int e^{2x}\cos(3x)dx}
5)
∫
e
−
x
(
2
x
2
−
3
x
+
5
)
d
x
{\displaystyle \int e^{-x}(2x^{2}-3x+5)dx}
6)
∫
4
tan
−
1
(
3
x
)
d
x
{\displaystyle \int 4\tan ^{-1}({3 \over x})dx}
7)
∫
x
tan
(
x
−
4
x
2
)
d
x
{\displaystyle \int x\tan(x-4x^{2})dx}
8)
∫
x
5
cos
−
1
(
3
x
)
d
x
{\displaystyle \int x^{5}\cos ^{-1}(3x)dx}
9)
∫
e
2
x
3
x
−
2
d
x
{\displaystyle \int e^{2x}3x^{-2}dx}
10)
∫
x
ln
(
3
x
)
d
x
{\displaystyle \int x\ln(3x)dx}