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These are the solutions to the Cambridge O-Level (Ordinary Level) past questions on Integral Calculus.
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$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int ax^n dx = \dfrac{ax^{n + 1}}{n + 1} + C \\[7ex] (2.)\:\: \displaystyle\int (ax \pm b)^n dx = \dfrac{(ax \pm b)^{n + 1}}{a(n + 1)} + C \\[7ex] (3.)\:\: \displaystyle\int (-ax \pm b)^n dx = -\dfrac{(ax \pm b)^{n + 1}}{a(n + 1)} + C \\[5ex] $
$ \text{a, b, k, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[7ex] (2.)\:\: \displaystyle\int e^x dx = e^x + C \\[7ex] (3.)\:\: \displaystyle\int e^{kx} dx = \dfrac{e^{kx}}{k} + C \\[7ex] (4.)\:\: \displaystyle\int e^{-kx} dx = \dfrac{-e^{-kx}}{k} + C \\[7ex] (5.)\:\: \displaystyle\int e^{ax \pm b} dx = \dfrac{e^{ax \pm b}}{n} + C \\[7ex] (6.)\:\: \displaystyle\int k^{ax \pm b} dx = \dfrac{k^{ax \pm b}}{a\ln k} + C \\[7ex] (7.)\:\: \displaystyle\int xe^{ax} dx = \dfrac{e^{ax}(ax - 1)}{a^2} + C \\[7ex] (8.)\:\: \displaystyle\int x^ne^{ax} dx = \dfrac{x^ne^{ax}}{a} - \dfrac{n}{a}\displaystyle\int x^{n - 1}e^{ax} dx \\[7ex] (9.)\:\: \displaystyle\int \dfrac{dx}{1 + ke^{ax}} = x - \dfrac{\ln(1 + ke^{ax})}{a} + C \\[5ex] $
$ \text{n is a positive constant} \\[3ex] (1.)\:\: \displaystyle\int \ln x dx = x\ln x - x + C \\[7ex] (2.)\:\: \displaystyle\int (\ln x)^n dx = x(\ln x)^n - n\displaystyle\int (\ln x)^{n - 1}dx \\[7ex] (3.)\:\: \displaystyle\int \dfrac{dx}{x\ln x} = \ln|\ln x| + C \\[7ex] (4.)\:\: \displaystyle\int x^n\ln x dx = \dfrac{x^{n + 1}[\ln x(n + 1) - 1]}{(n + 1)^2} + C \\[5ex] $
$ (1.)\:\: \displaystyle\int \sin x dx = -cos x + C \\[7ex] (2.)\:\: \displaystyle\int cos x dx = sin x + C \\[7ex] (3.)\:\: \displaystyle\int \sec^2 x = \tan x + C \\[5ex] $
$ (1.)\:\: \displaystyle\int \sinh x dx = \cosh x + C \\[7ex] (2.)\:\: \displaystyle\int \cosh x dx = \sinh x + C \\[5ex] $
$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int \dfrac{1}{x} dx = \ln x + C \\[7ex] (2.)\:\: \displaystyle\int \dfrac{1}{ax \pm b} dx = \dfrac{\ln|ax \pm b|}{a} + C \\[7ex] (3.)\:\: \displaystyle\int \dfrac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}x + C \\[7ex] (4.)\:\: \displaystyle\int \dfrac{-1}{\sqrt{1 - x^2}} dx = \cos^{-1}x + C \\[7ex] (5.)\:\: \displaystyle\int \dfrac{1}{1 + x^2} dx = \tan^{-1}x + C \\[7ex] (6.)\:\: \displaystyle\int \dfrac{1}{\sqrt{x^2 + 1}} dx = \sinh^{-1}x + C \\[7ex] (7.)\:\: \displaystyle\int \dfrac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}x + C \\[7ex] (8.)\:\: \displaystyle\int \dfrac{1}{1 - x^2} dx = \tanh^{-1}x + C \\[5ex] $
$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int |x| dx = \dfrac{x|x|}{2} + C \\[7ex] (2.)\:\: \displaystyle\int |ax \pm b| dx = \dfrac{(ax \pm b)|ax \pm b|}{2a} + C \\[5ex] $
$
\underline{\text{Algebraic Substitution}} \\[3ex]
(1.)\:\: \displaystyle\int f(x)f'(x) dx = \dfrac{f^2(x)}{2} + C \\[7ex]
(2.)\:\: \displaystyle\int \dfrac{f'(x)}{f(x)} dx = \ln f(x) + C \\[7ex]
(3.)\:\: \displaystyle\int \dfrac{-f'(x)}{f(x)} dx = -\ln f(x) + C \\[7ex]
\underline{\text{Trigonometric Substitution}} \\[3ex]
(4.)\;\; \displaystyle\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex]
(5.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex]
(6.)\;\; \displaystyle\int \sqrt{a^2 - x^2}dx = \dfrac{a^2}{2}\left[\sin^{-1}\left(\dfrac{x}{a}\right) +
\dfrac{x\sqrt{a^2 - x^2}}{a^2}\right] + C \\[7ex]
\underline{\text{Integration by Parts (Integration of Products)}} \\[3ex]
(7.)\;\; \displaystyle\int vdu = uv - \displaystyle\int udv \\[7ex]
\underline{\text{Hyperbolic Substitution}} \\[3ex]
(8.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{a^2 + x^2}} = \sinh^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex]
(9.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex]
\underline{\text{Integral Reflection Method}} \\[3ex]
\text{Also known as Symmetric Substitution OR Functional Transformation in Integration} \\[3ex]
\displaystyle\int_a^b \dfrac{f(x_1)}{f(x_1) + f(x_2)}dx \\[7ex]
Let\;\; a + b = \text{some constant},\;c \\[3ex]
\text{Set the substitution}:\;\; x = c - x \\[3ex]
\implies \\[3ex]
f(x) = \text{original function} \\[3ex]
f(c - x) = \text{transformed function} \\[3ex]
\text{Assume the original function and the transformed function complements each other} \\[3ex]
f(x) + f(c - x) = 1 \\[3ex]
Let\;\; I_1 = \displaystyle\int_a^b \dfrac{f(x_1)}{f(x_1) + f(x_2)}dx \quad\dots\text{integral of original
function}
\\[7ex]
Let:\;\;I_2 = \displaystyle\int_a^b \dfrac{f(x_2)}{f(x_2) + f(x_1)}dx \quad\dots\text{integral of transformed
function}
\\[7ex]
I_1 = I_2 = \text{the integral},\;I \\[4ex]
I_1 + I_2 = 2I \\[3ex]
$
Because we are integrating over the same limits and the variable, x is just being swapped with c
− x, the integral remains the same as long as the transformation preserves the total area under the
curve.
$
2I = \displaystyle\int_a^b 1 dx = [x]_a^b = b - a \\[5ex]
I = \dfrac{b - a}{2}
$