A function $F(x)$ is called an antiderivative of $f(x)$ on an interval if $F'(x) = f(x)$ for every $x$ in that interval. If $F(x)$ is one antiderivative, then every antiderivative has the form $F(x) + C$, where $C$ is an arbitrary constant.
The set of all antiderivatives is called the indefinite integral:$$\int f(x)\,dx = F(x) + C, \quad \text{where } F'(x) = f(x).$$
Key checking rule: integration and differentiation are inverse operations,$$\left(\int f(x)\,dx\right)' = f(x), \qquad \int F'(x)\,dx = F(x) + C.$$On the YÖS exam you can always verify an answer by differentiating it.
Memorize this table — almost every YÖS integral reduces to it:$$\int dx = x + C, \qquad \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \ (n \neq -1),$$$$\int \frac{dx}{x} = \ln|x| + C, \qquad \int e^x\,dx = e^x + C, \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C,$$$$\int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C,$$$$\int \sec^2 x\,dx = \tan x + C, \qquad \int \csc^2 x\,dx = -\cot x + C,$$$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C, \qquad \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C.$$
Quick examples: $\int 3x^8\,dx = 3\cdot\frac{x^9}{9} + C = \frac{x^9}{3} + C$; $\int \frac{dx}{\sqrt{4 - x^2}} = \arcsin\frac{x}{2} + C$ (here $a = 2$).
The integral is linear: constants factor out and the integral of a sum is the sum of the integrals,$$\int \big(\alpha f(x) + \beta g(x)\big)\,dx = \alpha \int f(x)\,dx + \beta \int g(x)\,dx.$$
This lets you integrate any polynomial term by term:$$\int (2x + a)\,dx = x^2 + ax + C.$$
Warning: there is no product or quotient rule for integrals — $\int f g\,dx \neq \int f\,dx \cdot \int g\,dx$. Products require substitution, integration by parts, or an algebraic/trigonometric transformation first.
If the integrand has the form $f(g(x))\,g'(x)$, substitute $u = g(x)$, $du = g'(x)\,dx$:$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.$$
How to spot it: one factor is a function of some expression, the other factor is (up to a constant) the derivative of that expression.
Typical YÖS patterns:- $\int \sin^4 x \cos x\,dx$: here $u = \sin x$, $du = \cos x\,dx$, so the integral becomes $\int u^4\,du = \frac{u^5}{5} + C = \frac{\sin^5 x}{5} + C$.- $\int e^{\tan x} \sec^2 x\,dx$: here $u = \tan x$, $du = \sec^2 x\,dx$, so we get $\int e^u\,du = e^{\tan x} + C$.- Linear substitution: $\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C$, e.g. $\int e^{2x}\,dx = \frac{e^{2x}}{2} + C$.
$$\int u\,dv = uv - \int v\,du.$$
Choose $u$ by the LIATE priority (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential): $u$ is the function that comes first in the list; the rest is $dv$.
Typical cases:- $\int x e^{2x}\,dx$, $\int x^2 \sin x\,dx$ — take $u$ = the polynomial; for $x^2$ apply the formula twice.- $\int \ln^2 x\,dx$ — take $u = \ln^2 x$, $dv = dx$; the result is $x\ln^2 x - 2(x\ln x - x) + C$, hence $\int_1^e \ln^2 x\,dx = e - 2$.- Fast trick for $\int e^x\big(f(x) + f'(x)\big)dx = e^x f(x) + C$. Example: $\int e^x(x^2 + x + 1)\,dx$: look for $f$ with $f + f' = x^2 + x + 1$, i.e. $f(x) = x^2 - x + 2$, so the answer is $e^x(x^2 - x + 2) + C$.
Two standard moves for $\int \frac{P(x)}{Q(x)}\,dx$:
1. If $\deg P \geq \deg Q$ — divide first.$$\int \frac{x^2}{x^2 + 1}\,dx = \int \left(1 - \frac{1}{x^2 + 1}\right)dx = x - \arctan x + C.$$
2. If the denominator factors — use partial fractions.$$\int \frac{5}{x^2 + x - 6}\,dx = \int \frac{5}{(x + 3)(x - 2)}\,dx = \int \left(\frac{1}{x - 2} - \frac{1}{x + 3}\right)dx = \ln\left|\frac{x - 2}{x + 3}\right| + C.$$The coefficients come from $\frac{5}{(x+3)(x-2)} = \frac{A}{x-2} + \frac{B}{x+3}$: multiply out and plug in $x = 2$ and $x = -3$ to get $A = 1$, $B = -1$.
Also remember the log pattern: $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.
Product to sum (for $\int \sin ax \cos bx\,dx$ etc.):$$\cos A \cos B = \tfrac{1}{2}\big[\cos(A - B) + \cos(A + B)\big], \quad \sin A \sin B = \tfrac{1}{2}\big[\cos(A - B) - \cos(A + B)\big],$$$$\sin A \cos B = \tfrac{1}{2}\big[\sin(A - B) + \sin(A + B)\big].$$Example: $\int \cos 5x \cos x\,dx = \tfrac{1}{2}\int (\cos 4x + \cos 6x)\,dx = \frac{\sin 4x}{8} + \frac{\sin 6x}{12} + C$.
Powers of $\tan x$: use $1 + \tan^2 x = \sec^2 x$.$$\int (\tan^4 x + \tan^2 x)\,dx = \int \tan^2 x\,(\tan^2 x + 1)\,dx = \int \tan^2 x \sec^2 x\,dx = \frac{\tan^3 x}{3} + C$$(substitution $u = \tan x$). Similarly $\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C$.
Even powers of $\sin, \cos$: lower the degree with $\sin^2 x = \frac{1 - \cos 2x}{2}$, $\cos^2 x = \frac{1 + \cos 2x}{2}$.
The definite integral $\int_a^b f(x)\,dx$ is a number (for $f \geq 0$ it equals the area under the graph from $a$ to $b$). It is computed by the Newton-Leibniz formula: if $F' = f$, then$$\int_a^b f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a).$$The constant $C$ cancels, so it is never written in definite integrals.
Examples:$$\int_2^4 \frac{dx}{x} = \ln x \Big|_2^4 = \ln 4 - \ln 2 = \ln 2;$$$$\int_0^3 (x - 3)^2\,dx = \frac{(x - 3)^3}{3}\Big|_0^3 = 0 - \frac{(-3)^3}{3} = 9.$$
Key properties:$$\int_a^a f\,dx = 0, \qquad \int_a^b f\,dx = -\int_b^a f\,dx, \qquad \int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx.$$
Symmetry on $[-a, a]$: if $f$ is odd, $\int_{-a}^{a} f(x)\,dx = 0$; if $f$ is even, $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$. Example: $\int_{-1}^{1} x^3 \cos x\,dx = 0$ (odd integrand) — no computation needed.
Piecewise functions: split the interval at the points where the formula changes. If $f(x) = x + 1$ for $x < 0$ and $f(x) = x^2 + 1$ for $x \geq 0$, then$$\int_{-1}^{1} f(x)\,dx = \int_{-1}^{0} (x + 1)\,dx + \int_0^1 (x^2 + 1)\,dx = \frac{1}{2} + \frac{4}{3} = \frac{11}{6}.$$The same splitting handles $|x|$: rewrite the absolute value piecewise first.
If $f$ is continuous, then$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x), \qquad \frac{d}{dx}\int_x^a f(t)\,dt = -f(x).$$
General rule with functions in the limits (chain rule):$$\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f\big(v(x)\big)\,v'(x) - f\big(u(x)\big)\,u'(x).$$
Important: you do not need to compute the integral itself — just plug the limit into $f$ and multiply by its derivative. This is exactly what YÖS problems like $f(x) = \int_x^3 (2t^2 - 3)\,dt$, find $f'(1)$, are testing.
Problem. Given $\int_{-2}^{0} (2x + a)\,dx = -2$, find $a$.
Solution.Step 1. Find the antiderivative: $\int (2x + a)\,dx = x^2 + ax$.Step 2. Apply Newton-Leibniz:$$\int_{-2}^{0} (2x + a)\,dx = \big(x^2 + ax\big)\Big|_{-2}^{0} = 0 - \big((-2)^2 + a(-2)\big) = -(4 - 2a) = 2a - 4.$$Step 3. Solve the equation: $2a - 4 = -2$, so $2a = 2$ and $a = 1$.
Answer: $a = 1$.
Problem. Compute $\int \sin^4 x \cos x\,dx$.
Solution.Step 1. Notice that $\cos x$ is the derivative of $\sin x$ — a perfect setup for substitution.Step 2. Let $u = \sin x$; then $du = \cos x\,dx$.Step 3. The integral becomes$$\int u^4\,du = \frac{u^5}{5} + C.$$Step 4. Return to $x$: the answer is $\frac{\sin^5 x}{5} + C$.
Check: $\left(\frac{\sin^5 x}{5}\right)' = \sin^4 x \cos x$. Correct.
The same idea gives $\int e^{\tan x}\sec^2 x\,dx = e^{\tan x} + C$ with $u = \tan x$, $du = \sec^2 x\,dx$.
Problem. Compute $\int x e^{2x}\,dx$.
Solution.Step 1. By LIATE, the algebraic factor is $u$: take $u = x$, $dv = e^{2x}\,dx$.Step 2. Then $du = dx$ and $v = \int e^{2x}\,dx = \frac{e^{2x}}{2}$.Step 3. Apply $\int u\,dv = uv - \int v\,du$:$$\int x e^{2x}\,dx = \frac{x e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}}{4}(2x - 1) + C.$$
Answer: $\frac{e^{2x}}{4}(2x - 1) + C$.
For $\int x^2 \sin x\,dx$ the same scheme is applied twice: first $u = x^2$, then $u = x$; the result is $-x^2\cos x + 2x\sin x + 2\cos x + C$.
Problem. Let $f(x) = \int_x^3 (2t^2 - 3)\,dt$. Find $f'(1)$.
Solution.Step 1. The variable $x$ is in the lower limit, so flip the limits and change the sign:$$f(x) = -\int_3^x (2t^2 - 3)\,dt.$$Step 2. By the fundamental theorem, $\frac{d}{dx}\int_3^x (2t^2 - 3)\,dt = 2x^2 - 3$, hence$$f'(x) = -(2x^2 - 3) = 3 - 2x^2.$$Step 3. Substitute $x = 1$: $f'(1) = 3 - 2 = 1$.
Answer: $f'(1) = 1$. Note that we never computed the integral itself — only substituted the limit into the integrand.
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