Indefinite integrals: standard, substitution, by parts

Mid-October 2026: in Year 5, at the liceo scientifico (science-oriented high school) and the other courses with the full maths programme, you are at derivatives and the theorems of Rolle, Lagrange and De L’Hôpital. Indefinite integrals usually start in November-December, and the first test on the topic falls six to ten weeks away, in December or January. The post is out before your teacher begins, on purpose: integrals are learned better if you arrive at the first lesson with a clear idea of what you are doing, and with the table of derivatives already in your head.

Integrals are the Year 5 topic where the difference between those who have a method and those who do not shows at once. There is no single algorithm as there is for derivatives: you have to choose the right technique. But the techniques are few, and the choice is made with a couple of questions. Here you will find the minimum theory, the table, the three techniques of the high-school syllabus and three worked exercises, with the check done by differentiating the result.

Antiderivative and indefinite integral

A function $F(x)$ is an antiderivative of $f(x)$ if its derivative is $f(x)$:

If $F$ is an antiderivative, so is $F+3$, and $F-100$, because the derivative of a constant is zero. That is why the set of all antiderivatives is written with an arbitrary constant, and it is the indefinite integral:

Integrating is therefore the inverse operation of differentiating. And that is the reason why the check is always possible: you differentiate the result and you must find the function you were integrating. If you have studied derivatives well, integrals are the same table read backwards. If instead you still have doubts about derivatives, sort them out now: the first month of integrals is full of them.

The table of standard integrals

These are the results to know by heart. They are the derivatives you know, written backwards.

Two notes. The absolute value in $\ln |x|$ is not a detail: it is needed because $\frac{1}{x}$ is defined for $x<0$ too. And the case $n=-1$ is the exception to the first row, and it is the one that produces the logarithm.

Linearity

Two properties that you use in every exercise: the constant comes out of the integral, and the integral of a sum is the sum of the integrals.

That is why a polynomial is integrated term by term:

Beware, though: the same thing does not hold for a product. $\int x\cdot e^{x}dx$ is not $\int xdx\cdot \int e^{x}dx$. It is the typical mistake of someone studying in a hurry, and for products you need the techniques below.

Substitution

The idea is the inverse of the chain rule. If inside the integral there is a composite function multiplied by the derivative of the inner function, you set $t=g(x)$ and the integral becomes simpler. The structure you must learn to recognise is

Three cases that always appear and that are worth knowing as “quick rules”:

When the inner derivative is missing by a constant, you adjust for it. For example

because the derivative of $3x$ is $3$, and I have to divide by $3$ to compensate. Forgetting that $\frac{1}{3}$ is one of the mistakes I see most: the check flags it at once, because differentiating $\sin (3x)$ gives $3\cos (3x)$.

If the inner derivative is missing by a function of $x$ (not a constant), the method usually does not work and another route has to be found.

Integration by parts

From the product rule for differentiation we get

You use it when there is a product of functions of different types, for example a polynomial times an exponential, or a logarithm on its own. The choice of which is $u$ (the part you differentiate) and which is $v^{\prime }$ (the part you integrate) decides whether the exercise becomes easy or impossible.

The mnemonic is LIATE, the order of priority for $u$:

• Logarithms
• Inverse trigonometric (arcsin, arctan)
• Algebraic (polynomials, powers)
• Trigonometric
• Exponentials

The first family in the list that appears in the integral is $u$; the rest, with $dx$, is $v^{\prime }$. It is not a law of nature, it is a guide that almost always works in high school. The real criterion is: $u$ must get simpler when you differentiate it, and $v^{\prime }$ must be easy to integrate.

Example:

Algebraic and trigonometric: $u=x$, $v^{\prime }=\cos x$, so $u^{\prime }=1$ and $v=\sin x$. Then

If I had chosen the other way round ($u=\cos x$, $v^{\prime }=x$) the integral would have got more complicated, with $x^2$ appearing in the new integral. That is the signal that you have swapped the roles. Sometimes integration by parts has to be applied twice in a row (for example with $x^2{e}^x$): each step lowers the degree of the polynomial by one, until it disappears.

Rational functions

These are integrals of the type $\frac{N(x)}{D(x)}$, polynomial over polynomial. In high school the denominator usually has degree $2$ and the numerator a lower degree (if the numerator has equal or higher degree, you first do polynomial division). The procedure depends on the discriminant $\Delta$ of the denominator.

Case $\Delta >0$: the denominator has two distinct roots $x_1,x_2$ and factorises as $(x-x_1)(x-x_2)$. The fraction is split into partial fractions:

The numbers $A$ and $B$ are found by substituting $x=x_1$ and $x=x_2$ into the identity, after clearing the denominators. Each of the two fractions then gives a logarithm.

Case $\Delta =0$: the denominator is a square, $(x-x_0{)}^2$. It is convenient to rewrite the numerator in terms of $x-x_0$, so as to obtain a logarithm plus a negative power:

Case $\Delta <0$: the denominator does not factorise and the arctangent appears, often together with a logarithm. In many high-school syllabuses it is seen only in its simple form, such as $\int \frac{1}{x^2+1}dx$, so I do not cover it here: ask your teacher how far they go. If you want to see how that case is reached too, the university version of these topics is in integrals explained well for Calculus 1, where the approach is more formal and includes improper integrals.

Three worked exercises

Try to solve them before looking. In all three, at the end, I differentiate the result to check it.

Exercise 1: substitution

Inside there is $x^2+3$ and outside $x$, which is half of its derivative ($2x$). I set $t=x^2+3$, so $dt=2xdx$ and $xdx=\frac{1}{2}dt$:

Forgetting to go back to $x$ is the classic mistake: the answer must contain $x$, not $t$.

Check: I differentiate $\frac{(x^2+3{)}^5}{10}$ with the chain rule:

It is the starting function. It works.

Exercise 2: by parts

By LIATE the logarithm comes before the algebraic function: $u=\ln x$, $v^{\prime }=x$. Then $u^{\prime }=\frac{1}{x}$ and $v=\frac{x^2}{2}$:

The last integral is a standard one:

Check: I differentiate the result.

It works. To train with tests in the style of the Year 5 exam there are the maturità maths practice papers: they are designed for the second paper.

Exercise 3: rational function with $\Delta >0$

The denominator has $\Delta =25-24=1>0$ and factorises as $(x-2)(x-3)$. I look for $A$ and $B$ such that

I clear the denominators:

I set $x=2$: $3=A(-1)$, so $A=-3$. I set $x=3$: $4=B\cdot 1$, so $B=4$. The integral becomes

Check: I differentiate the result and put it over a common denominator.

It works. If I had had $\Delta =0$, for example $\int \frac{x}{(x-1{)}^2}dx$, I would have used the rewriting seen above and obtained $\ln |x-1|-\frac{1}{x-1}+c$.

How to choose the technique

Faced with an integral, do not start at random. Ask yourself these questions, in this order:

1. Is it a standard one? Check the table, even after adjusting a constant (as in $\cos (3x)$).
2. Is there a composite function with its derivative? Then it is a substitution, or one of the three quick rules: $f^{\prime }{f}^n$, $\frac{f^{\prime }}{f}$, $f^{\prime }{e}^f$.
3. Is it a product of functions from different families? Then it is by parts, with LIATE to choose $u$.
4. Is it a rational function? Then you look at the $\Delta$ of the denominator and use partial fractions.

If none works, before giving up try rewriting the integrand: splitting a fraction, dividing, completing the square. Many “difficult” integrals become standard after a two-line rewrite.

The mistakes that cost marks

• Forgetting the $+c$. In an indefinite integral it is compulsory, and many teachers take off half a mark or a whole mark each time.
• Forgetting the inner derivative. $\int e^{2x}dx$ is not $e^{2x}$ but $\frac{1}{2}{e}^{2x}$. The check flags it at once.
• Splitting a product as $\int fg=\int f\cdot \int g$. It does not exist, ever.
• Swapping $u$ and $v^{\prime }$ in the by-parts method and ending up with an integral harder than the starting one. If it gets worse, you chose badly.
• Forgetting the absolute value in the logarithm: $\ln |x|$, not $\ln x$, unless the text says $x>0$.
• Not going back to $x$ after a substitution: the final result must not contain $t$.
• Signs in the derivatives of trigonometric functions: $\int \sin xdx=-\cos x$, with the minus. If you lose it, the check does not work.
• Not checking. Differentiating the result costs a minute and flags almost all mistakes, because differentiation is a mechanical procedure you already know well.

How to prepare over the coming weeks

The test is far off, and that is an advantage: integrals are learned by volume, not by intuition. The plan I recommend:

1. Now, before your teacher starts: review the table of derivatives, in particular the chain rule. If you need to review the theorems, Rolle, Lagrange and De L’Hôpital with exercises rebuilds them one by one, and function study with the first derivative is a good test for derivatives.
2. In the first two weeks of integrals: learn the table by heart and do at least ten standard ones a day, for the first few days. They are the base on which everything else rests.
3. Then substitution and by parts: five exercises per type, always checked by differentiating. The check is your own marking: if you can do it, you do not need anyone to tell you whether you have made a mistake.
4. Two weeks before the test: mix the types. The test does not tell you which technique to use, so practising recognising it is half the training.

A note: in Year 5 integrals do not end with indefinite ones. Next come the definite integral, areas and volumes, and that is the part that appears in the second paper of the maturità (Italy’s school-leaving exam). Everything you do on indefinite integrals is needed there, so there is no point settling for a barely passing grade now. The general overview of the subject is on the Maths page.

If you cannot choose the technique

The difficulty of almost all Year 5 students, on integrals, is not the calculation: it is the first glance, the one that should tell you “this is by parts” or “there is a substitution here”. It is trained with plenty of examples looked at together, aloud. If you want help, write to me from the contact page: we do a free introductory call, you send me three integrals that do not work out for you, and I tell you where the choice of method jams.