Integration by Parts (IBP) comes from the product rule: d(uv) = u dv + v du, giving integral u dv = uv - integral v du. The challenge is choosing u and dv wisely.

LIATE Rule: Choose u from this priority: Logarithmic (ln x, log x), Inverse trig (arctan x, arcsin x), Algebraic ($x^n$, polynomials), Trigonometric (sin x, cos x), Exponential ($e^x$, $a^x$). The function with higher priority becomes u.

Tabular Method (Repeated By-Parts): For integral $x^n$ * e^(ax) dx or integral $x^n$ * sin(ax) dx, create two columns: derivatives of the algebraic part and integrals of the other part. Alternate signs (+, -, +, -...) and multiply diagonally until the algebraic part reaches zero.

Example: integral $x^3$ * $e^x$ dx

Result: $e^x$($x^3$ - 3$x^2$ + 6x - 6) + C

Cyclic By-Parts: For integral e^(ax)*sin(bx) dx or integral e^(ax)*cos(bx) dx, applying IBP twice brings back the original integral. Let I denote the original integral, then solve algebraically:

• I = e^(ax)(asin(bx) - b*cos(bx))/(a^{2+b}^2) + C
• integral e^(ax)cos(bx) dx = e^(ax)(acos(bx) + bsin(bx))/(a^{2+b}^2) + C

The $e^x$ Shortcut: integral $e^x$[f(x) + f'(x)] dx = $e^x$*f(x) + C. This is the most tested IBP-related result in JEE. To apply: (1) Check if the non-exponential part can be split as g(x) + g'(x), (2) If yes, the answer is $e^x$*g(x) + C.