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Indefinite Integration JEE Brief

This lecture covers core concepts and problem-solving techniques for indefinite integration targeted at JEE Mains and Advanced. Topics include standard formulas, algebraic substitutions, integration by parts (ILATE rule and DI shortcut), standard quadratic forms, and partial fractions.

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· 8 topics
Checked against the video on 17 Oct 2026

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Indefinite Integration JEE Brief

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This lecture covers core concepts and problem-solving techniques for indefinite integration targeted at JEE Mains and Advanced. Topics include standard formulas, algebraic substitutions, integration by parts (ILATE rule and DI shortcut), standard quadratic forms, and partial fractions.

Introduction and Basic Formulas

Integration is fundamentally the reverse process of differentiation.
An arbitrary integration constant C is always added to indefinite integrals because the derivative of any constant is zero.
The standard power formula integral(xⁿ dx) = (xⁿ⁺¹)/(n+1) + C holds for all values of n except n = -1.
When n = -1, the integral integral(1/x dx) evaluates to ln|x| + C.
For exponential functions, integral(a^x dx) = a^x / ln(a) + C where a > 0 and a != 1.
Must knowStandard trigonometric integrals include integral(sin(x) dx) = -cos(x) + C, integral(cos(x) dx) = sin(x) + C, and integral(sec²(x) dx) = tan(x) + C.

Linear Substitution and Rules of Integration

Constants can be factored out: integral(k × f(x) dx) = k × integral(f(x) dx).
Integrals distribute over addition and subtraction: integral(f(x) ± g(x) dx) = integral(f(x) dx) ± integral(g(x) dx).
Must knowWhen a linear expression ax + b replaces x in any standard function f(x), evaluate the standard integral and divide by a (the coefficient of x).
Treating an inner expression as a linear term only works for linear functions ax + b, never for non-linear polynomials like quadratic expressions.

Method of Substitution

The four primary techniques to solve integrals are substitution, by parts (product rule), partial fractions, and miscellaneous algebraic/trigonometric methods.
Substitution involves setting a suitable expression to t such that its derivative appears alongside dx as dt.
Must knowStandard integrals derived via substitution: integral(tan(x) dx) = ln|sec(x)| + C and integral(cot(x) dx) = ln|sin(x)| + C.
Must knowThe integral of secant is integral(sec(x) dx) = ln|sec(x) + tan(x)| + C or ln|tan(pi/4 + x/2)| + C.
The integral of cosecant is integral(csc(x) dx) = ln|csc(x) - cot(x)| + C or ln|tan(x/2)| + C.

Repeated JEE Advanced and Mains Algebraic Patterns

For integrals of the form integral(dx / ((x-a)^p × (x-b)^q)) where powers sum to an integer, divide and multiply by a factor to create a common ratio power like ((x-a)/(x-b))^k.
Substituting the ratio ((x-a)/(x-b)) = t reduces the remaining denominator factor and dx into a simple dt expression.
For expressions with large powers inside brackets, factoring out the highest power of x produces negative powers that match the numerator derivative.
Taking x⁴ out of a bracket raised to power 4 yields x¹⁶ outside, which divides the numerator into terms matching the derivative of the inner bracket.

General Trigonometric Substitutions and 7 Standard Quadratic Formulas

Expressions containing sqrt(a² - x²) require the substitution x = a*sin(theta).
Expressions containing sqrt(a² + x²) use x = a*tan(theta), while sqrt(x² - a²) uses x = a*sec(theta).
Integrals containing sqrt((a-x)/(a+x)) simplify using x = a*cos(2*theta) to eliminate the radical via half-angle identities.
Must knowintegral(dx / (x² + a²)) = (1/a) × tan⁻¹(x/a) + C.
Must knowintegral(dx / sqrt(a² - x²)) = sin⁻¹(x/a) + C (note: no 1/a factor appears in front of sin⁻¹).
Must knowintegral(dx / (x² - a²)) = (1/(2a)) × ln|(x-a)/(x+a)| + C, while integral(dx / (a² - x²)) = (1/(2a)) × ln|(a+x)/(a-x)| + C.
For integrals of the form integral(dx / sqrt(x² ± a²)), the result is ln|x + sqrt(x² ± a²)| + C.

Linear over Quadratic and Completing the Square

Any integral of the form integral(dx / (ax² + bx + c)) or integral(dx / sqrt(ax² + bx + c)) is solved by completing the square on the quadratic term.
Completing the square transforms the quadratic into (x + b/(2a))² ± k², directly matching standard inverse trigonometric or logarithmic formulas.
Integrals of the form integral((px + q) / (ax² + bx + c) dx) are split by writing px + q = A × d/dx(ax² + bx + c) + B.
This split yields two parts: one directly solvable via substitution where the numerator is the exact derivative of the quadratic, and the second solvable by completing the square.

Integration by Parts and the DI Shortcut Method

Integration by parts follows the formula: integral(u × v dx) = u × integral(v dx) - integral((du/dx × integral(v dx)) dx).
The priority for choosing the first function (u) follows the ILATE rule: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.
In repeated integration by parts, the DI method organizes terms into Differentiation (D) and Integration (I) columns.
Alternating signs (+, -, +, -) are assigned to the D column starting with +, and terms are multiplied diagonally to form the integral without writing intermediate steps.
Must knowSpecial form: integral(e^x × (f(x) + f'(x)) dx) = e^x × f(x) + C.
Must knowSpecial form: integral(f(x) + x × f'(x) dx) = x × f(x) + C.

Partial Fractions

Rational expressions P(x)/Q(x) where the degree of P(x) is less than Q(x) can be decomposed into partial fractions.
For non-repeated linear factors (x - a)(x - b), write A/(x - a) + B/(x - b).
For repeated linear factors (x - a)³, the decomposition is A/(x - a) + B/(x - a)² + C/(x - a)³.
For non-factorable quadratic factors (x² + a), the numerator must be assigned a linear term: (Ax + B)/(x² + a).
For simple products of two linear factors in the denominator, partial fraction decomposition can be bypassed by expressing the constant numerator as the difference of the two denominator factors.

Key terms

Integration: The reverse process of differentiation
ILATE Rule: Priority order for choosing the first function in integration by parts: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential
DI Method: Tabular shortcut method for repeated integration by parts using differentiation and integration columns with alternating signs
Partial Fractions: Technique of decomposing a complex rational algebraic fraction into a sum of simpler fractions

Formulas

integral(xⁿ dx) = (xⁿ⁺¹)/(n+1) + C: Valid for n != -1
integral(tan(x) dx) = ln: sec(x)| + C | Integral of tangent function
integral(sec(x) dx) = ln: sec(x) + tan(x)| + C = ln|tan(pi/4 + x/2)| + C | Integral of secant function
integral(dx / (x² + a²)) = (1/a) × tan⁻¹(x/a) + C: Standard form yielding inverse tangent
integral(dx / sqrt(a² - x²)) = sin⁻¹(x/a) + C: Standard form yielding inverse sine
integral(dx / (x² - a²)) = (1/(2a)) × ln: (x - a)/(x + a)| + C | Standard form for difference of squares in denominator
integral(e^x × (f(x) + f'(x)) dx) = e^x × f(x) + C: Standard exponential product form
integral(f(x) + x × f'(x) dx) = x × f(x) + C: Standard product derivative form

Exam tips

When integrating 1/linear, always include absolute value bars: ln|ax + b| / a + C to prevent errors with domain restrictions.
In the standard formula integral(dx / sqrt(a² - x²)) = sin⁻¹(x/a) + C, do not multiply by 1/a; only tan⁻¹ and sec⁻¹ standard formulas have 1/a outside.

Quick revision

integral(1/x dx) = ln|x| + C for x != 0.
When a linear expression ax + b replaces x in any formula, divide the whole integral by a.
integral(sec(x) dx) equals ln|sec(x) + tan(x)| + C or ln|tan(pi/4 + x/2)| + C.
integral(dx / (x² - a²)) = (1/(2a)) × ln|(x-a)/(x+a)| + C.
integral(dx / (a² - x²)) = (1/(2a)) × ln|(a+x)/(a-x)| + C.
integral(dx / sqrt(x² + a²)) = ln|x + sqrt(x² + a²)| + C.
integral(dx / sqrt(x² - a²)) = ln|x + sqrt(x² - a²)| + C.
The expression integral(e^x × (f(x) + f'(x)) dx) evaluates directly to e^x × f(x) + C.
The expression integral(f(x) + x × f'(x) dx) evaluates directly to x × f(x) + C.
For integrals with sqrt(a² - x²), substitute x = a*sin(theta).

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