0Integration by parts can be applied to certain problems which are in the form of:
∫ f(x) g(x) dx
Put either one as 'u' and other one as 'dv' then solve using the formula
∫ u dv = uv - ∫ v du-------------------------------A
While choosing the function ' u ' choose that function which gets simpler by differentiation. As a thumb rule, a function to be chosen as ' u ' has to be in the following order (Logarithmic function, Inverse trig functions, Algebraic functions, Trig functions, Exponential functions or memorise L I A T E)
For example, in a function ∫ x² e× dx , e× is exponential function and x² is a algebraic function. Suppose we choose e× as ' u ' and x² as 'dv' , can we get a solution with ease ? Let's try one example.
∫ x² e× dx ( let u = e×, then du = e× dx and let dv = x² then v = ∫ x² dx = x³ / 3
substituting u & dv in the formula "A"
∫ e× x² dx = e× x³ / 3 - 1/3 ∫ x³ e× dx ------------------B
Again let u = e×, then du = e× dx and let dv = x³ then v = ∫ x³ dx = x^4 /4 substituting once again u & dv in "B"
∫ e× x² dx = e× x³ / 3 - 1/3 [ { e× x^4 / 4 - ∫ x^4 e× dx } ] and it goes on and on.
Why ? Because we selected u= e×, and dv= x², means we chose "u" for Exponential function and "dv" for an Algebraic function.
Let's now follow rule L I A T E and try to solve this problem
Let u = x², then du = 2 x dx and let dv = e× then v = ∫ e× dx = e×.
By substituting u & dv
∫ x² e× dx = x² e× - ∫ e× 2 x dx
= x² e× - 2 ∫ x e× dx, again let u = x then du = 1 dx & let dv = e× then v = e×
= x² e× - 2 { x e× - ( ∫ e× dx ) }
= x² e× - 2 { x e× - e× } + c
= x² e× - 2 x e× + 2 e× + c
= e× ( x² - 2 x + 2 ) + c
∴ ∫ x² e× dx = e× ( x² - 2 x + 2 )
