Integration is the antiderivatives in calculus.
Integration can be divided into two parts,
(a) Indefinite integral and
(b) definite integral.
In the case of indefinite integrals we have an arbitrary constant whose value is not know. whereas in the case of definite integrals, we evaluate the integral between the two points either on the x-axis or y-axis, which are called the lower limits and the upper limits of the integral. The value of the integral is definite which is the area bounded by the two curves between the two limits with the x- axis or y- axis.

We will be able to evaluate the integral only if we are comfortable with the different types of integrals. Integration by parts is the method of evaluating the integrals when we have product of two functions. In this section let us see about the procedure to follow the integration by parts and some of the examples.

When we have product of functions like, algebraic, trigonometric, logarithmic, exponential or inverse trigonometric functions, we can use the method of integration by parts.
The formula for integration by parts is given by,
$\int$ u dv = u . v - $\int$ v . du,
where u and v are the functions of x given by u = u (x) and v = v (x)

How to do Integration by Parts:

1. First we identify the u and dv of the given integral.

2. Differentiate u = u (x) with respect to x, to find $\frac{du}{dx}$.
Cross multiply and find du as a function of x, including dx.

3. Inegrate dv with respect to x, to get the function v = v (x).

4. Substitute for u, v, and du in the above formula.
$\int$ u dv = u . v - $\int$ v . du, Integrate the integral on the right side using the table of integration or substitution.

5. If the integral on the right side is the product of two functions, of the form, $\int$ u dv, continue the steps from 1 to 4, until we get a step without any integral sign.
Integration by Parts Theorem: If u and v are two functions of x, then,
$\int$ u v dx = u .($\int$ v dx) - $\int$ {$\frac{du}{dx}$ $\int$ v . dx } dx

(i. e) The integral of the product of two functions = (First Function) x (Integral of Second Function) - Integral of [(differential of first function) x (integral of second function)]

Proof: For any two functions f (x) and g (x) , we have,
$\frac{d}{dx}$ [f (x) . g (x)] = f (x) . $\frac{d}{dx}$ [g (x)] + g (x) . $\frac{d}{dx}$ [f (x)]

$\int$ [f (x). $\frac{d}{dx}$ [g (x)] + g (x) . $\frac{d}{dx}$ [f (x)]] dx = f (x) . g (x)

=> $\int$ [f (x) . $\frac{d}{dx}$ [g (x)] dx + $\int$ [g (x) . $\frac{d}{dx}$ [f (x)] dx = f (x) . g (x)

=> $\int$ [f (x) . $\frac{d}{dx}$ [g (x)]] = f (x) . g (x) - $\int$ [g (x) . $\frac{d}{dx}$ [f (x] ] dx
Substituting f (x) = u and $\frac{d}{dx}$ [g (x)] = v, so that g (x) = $\int$ v. dx,
we get,
$\int$ u . v dx = u .($\int$ v dx) - $\int$ {$\frac{du}{dx}$ $\int$ v . dx } dx

Solved Examples

Question 1: Evaluate $\int$ x.log x . dx
Solution:
 
We have, $\int$ x.log x . dx
Let us assume u = log x and  d v = x. dx
                      Since u = log x,
we have                   $\frac{du}{dx}$ = $\frac{1}{x}$

        =>                 d u = $\frac{1}{x}$. dx

                    Since d v = x. dx
                    we have v = $\int$ x. dx
        =>                    v = $\frac{x^{2}}{2}$,
Substituting u, v and du in the above formula for Integration by Parts, we get,
              $\int$ u dv = u . v  - $\int$ v . du
=>         $\int$ x.log x . dx =  log x . $\frac{x^{2}}{2}$ - $\int$ [ $\frac{x^{2}}{2}$ . $\frac{1}{x}$ ]. dx

                               = log x . $\frac{x^{2}}{2}$ - $\frac{1}{2}$ $\int$ [ x . dx ]

                               = $\frac{x^{2}\;log\;x\;}{2}$ - $\frac{1}{2}$ . $\frac{x^{2}}{2}$

  $\int$ x.log x . dx = $\frac{x^{2}\;log\;x\;}{2}$$\frac{x^{2}}{4}$ + C
 

Question 2: Integrate $\int$ arcsin x . dx
Solution:
 
We have $\int$ arcsin x . dx
The given integral is of the form, $\int$ u . dv
                        Substituting  u = arc sin x and dv = dx

We get              $\frac{du}{dx}$ = $\frac{1}{\sqrt{1-x^{2}}}$  [ Differentiating on both sides ]

                          =>           du =  $\frac{1}{\sqrt{1-x^{2}}}$ . dx

                                Since dv = dx

                                $\int$ dv = $\int$ dx

                                   =>    v = x
Substituting for, u, v and du in the formula for integration by parts,
                               $\int$ u dv = u . v  - $\int$ v . du
We get,      $\int$ arc sin x . dx = arc sin x . x - $\int$ x . $\frac{1}{\sqrt{1-x^{2}}}$ . dx

=>                                           = x . arc sinx - $\int$  $\frac{x}{\sqrt{1-x^{2}}}$ . dx

                                               = x . arc sin x - $\int$ $\frac{1}{t}$ ( - t dt )  [ substituting 1 - x2 = t2 ,
                                                                                                   we get     - 2x dx = 2 t d t
                                                                                                          =>      x dx  = - t dt ( dividing both sides by -2 ) ]
                                              = x arc sin x  + $\int$ 1. dt

                                             = x arc sin x + t

            $\int$ src sin x . dx  = x . arc sin x + $\sqrt{1-x^{2}}$ + C
 

Evaluate the following integrals.

Practice Problems

Question 1: $\int$ x. sin 3x dx
Question 2: $\int$ arc tan x dx
Question 3: $\int$ x log x . dx
Question 4: $\int$ x2 . ln 2x . dx.
Question 5: $\int$ x3 . ex . dx