A brief note on Integration as the Reverse of Differentiation . Read it before a test. Try each question at the end before you open its answer.
Antiderivatives Definition: F F F is an antiderivative of f f f when F ′ = f F' = f F ′ = f . The whole family is ∫ f ( x ) d x = F ( x ) + C \int f(x)\,dx = F(x) + C ∫ f ( x ) d x = F ( x ) + C . The constant: two antiderivatives differ by a constant. A point on the curve fixes C C C . Power rule: ∫ x n d x = x n + 1 n + 1 + C \int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C ∫ x n d x = n + 1 x n + 1 + C for n ≠ − 1 n \ne -1 n = − 1 , and ∫ d x x = ln ∣ x ∣ + C \int \dfrac{dx}{x} = \ln|x| + C ∫ x d x = ln ∣ x ∣ + C . Rules: constants and sums come out term by term. There is no rule for a product. Substitution Method: put u = g ( x ) u = g(x) u = g ( x ) and d u = g ′ ( x ) d x du = g'(x)\,dx d u = g ′ ( x ) d x . Replace every x x x , integrate, then return to x x x . Linear inside: ∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C \int f(ax + b)\,dx = \dfrac{1}{a}F(ax + b) + C ∫ f ( a x + b ) d x = a 1 F ( a x + b ) + C . Top is the derivative of the bottom: ∫ f ′ f d x = ln ∣ f ∣ + C \int \dfrac{f'}{f}\,dx = \ln|f| + C ∫ f f ′ d x = ln ∣ f ∣ + C . Four to know: ∫ tan x d x = ln ∣ sec x ∣ \int \tan x\,dx = \ln|\sec x| ∫ tan x d x = ln ∣ sec x ∣ , ∫ cot x d x = ln ∣ sin x ∣ \int \cot x\,dx = \ln|\sin x| ∫ cot x d x = ln ∣ sin x ∣ , ∫ sec x d x = ln ∣ sec x + tan x ∣ \int \sec x\,dx = \ln|\sec x + \tan x| ∫ sec x d x = ln ∣ sec x + tan x ∣ , ∫ cosec x d x = ln ∣ cosec x − cot x ∣ \int \operatorname{cosec} x\,dx = \ln|\operatorname{cosec} x - \cot x| ∫ cosec x d x = ln ∣ cosec x − cot x ∣ . Trigonometric identities Even powers: sin 2 x = 1 − cos 2 x 2 \sin^2 x = \dfrac{1 - \cos 2x}{2} sin 2 x = 2 1 − cos 2 x and cos 2 x = 1 + cos 2 x 2 \cos^2 x = \dfrac{1 + \cos 2x}{2} cos 2 x = 2 1 + cos 2 x . Odd powers: keep one factor for d u du d u and use sin 2 x + cos 2 x = 1 \sin^2 x + \cos^2 x = 1 sin 2 x + cos 2 x = 1 . Products: 2 sin A cos B = sin ( A + B ) + sin ( A − B ) 2\sin A\cos B = \sin(A + B) + \sin(A - B) 2 sin A cos B = sin ( A + B ) + sin ( A − B ) . Tangent: tan 2 x = sec 2 x − 1 \tan^2 x = \sec^2 x - 1 tan 2 x = sec 2 x − 1 . A worked example Find ∫ 2 x + 3 x 2 + 3 x + 1 d x \displaystyle\int \dfrac{2x + 3}{x^2 + 3x + 1}\,dx ∫ x 2 + 3 x + 1 2 x + 3 d x .
Look at the top. The derivative of the bottom x 2 + 3 x + 1 x^2 + 3x + 1 x 2 + 3 x + 1 is 2 x + 3 2x + 3 2 x + 3 , which is the top. Use f ′ / f f'/f f ′ / f . The integral is ln ∣ x 2 + 3 x + 1 ∣ + C \ln|x^2 + 3x + 1| + C ln ∣ x 2 + 3 x + 1∣ + C . Practice with new numbers
A program to try The program checks that F ( x ) = x 3 3 − x 2 + 4 F(x) = \dfrac{x^3}{3} - x^2 + 4 F ( x ) = 3 x 3 − x 2 + 4 has the slope x 2 − 2 x x^2 - 2x x 2 − 2 x at four points.
Run it. Do the two columns agree?
1. Change + 4 in F to + 9. Does anything change in the slopes? Why not?
Mistakes to avoid Forgetting + C + C + C . The answer is a family, not one function. Forgetting the factor of the inside. ∫ cos 5 x d x \int \cos 5x\,dx ∫ cos 5 x d x needs 1 5 \dfrac{1}{5} 5 1 . Mixing x x x and u u u . Return to x x x at the end. A wrong sign. ∫ sin x d x = − cos x \int \sin x\,dx = -\cos x ∫ sin x d x = − cos x . Check yourself Find ∫ ( 6 x 2 − 4 ) d x \displaystyle\int (6x^2 - 4)\,dx ∫ ( 6 x 2 − 4 ) d x . Stuck? Show a hint Show the answerFind ∫ x x 2 + 4 d x \displaystyle\int \dfrac{x}{x^2 + 4}\,dx ∫ x 2 + 4 x d x . Stuck? Show a hint Show the answerFind ∫ sin 3 x d x \displaystyle\int \sin 3x\,dx ∫ sin 3 x d x . Stuck? Show a hint Show the answerWrite cos 2 x \cos^2 x cos 2 x without a square. Stuck? Show a hint Show the answerA curve has the slope 3 x 2 3x^2 3 x 2 and passes through ( 1 , 5 ) (1, 5) ( 1 , 5 ) . Find its equation. Stuck? Show a hint Show the answerSources OpenStax, Calculus Volume 1 , Chapter 5, sections 5.4 and 5.5. NCERT, Mathematics Part II for Class XII, Chapter 7, the sections on integration as the inverse process of differentiation and on substitution.