This document discusses using integration by parts to evaluate integrals involving the product of two functions. Specifically, it shows how to integrate xsinx and integrals involving natural logarithms, such as ln x, by letting one function be u and the other dv, then applying the integration by parts formula.

1. H. Dobroshtan
2017

2. Suppose we wish to integrate the product of two functions,
such as xsinx, where one of the functions is not related to the
derivative of the other.
An expression such as this can be integrated using the method
of integration by parts.
udv uv vdu−∫ ∫=

3. So, to integrate xsinx with respect to x:
Let and=u x xdxdv = sin
So dx×du =1
We don’t need
the “+ c” here.Now, using the formula udv uv vdu−∫ ∫=
x xdx x x x dx− − −∫ ∫sin = cos ( cos )
x x x c− − −= cos ( sin )+
= sin cos +x x x c−
differentiate integrate
= cosv x−and

4. So anddx×du =1
2
=
2
x
e
v
Now, apply the formula udv uv vdu−∫ ∫=
2 2
2
=
2 2
x x
x e e
xe dx x dx−∫ ∫
Find .2x
xe dx∫
2 2
= +
2 4
x x
xe e
c−
2
= (2 1)+
4
x
e
x c−
x
e dx2
dv =Let and=u x

5. let and= lnu x = 8
dv
x
dx
so anddx
x
×
1
du = 2
= 4v x
Now, using the formula, udv uv vdu−∫ ∫=
2 2 1
8 ln = 4 ln 4 ×x xdx x x x dx
x
−∫ ∫
Find .8 lnx xdx∫
We don’t know the integral of lnx so:
2
= 4 ln 4x x xdx− ∫
2 2
= 4 ln 2 +x x x c−
2
= 2 (2ln 1)+x x c−

6. We can also use integration by parts to find the integral of ln x.
We do this by writing lnx as (1 × lnx).
Let and= lnu x dv = dx
So
dx
x
du =
Now we can integrate by parts:
1
1×ln = lnxdx x x x dx
x
−∫ ∫
= ln 1x x dx− ∫
= ln +x x x c−
ln = ln +xdx x x x c−∫
=v xand

7. We can also use integration by parts to find the integral of ln x.
We do this by writing lnx as (1 × lnx).
Let and= lnu x dv = dx
So
dx
x
du =
Now we can integrate by parts:
1
1×ln = lnxdx x x x dx
x
−∫ ∫
= ln 1x x dx− ∫
= ln +x x x c−
ln = ln +xdx x x x c−∫
=v xand