How to self-learn integrals using differentiation (but in reverse)?

Question

i'm having a hard time solving even basic and easy intergral questions highschoolers solve. lets say my question is intergral $(ax+b)^2$ then i think of "what could have been the question" that is $(ax+b)^3 $ then further differentiate it $(ax+b)^3=(ax+b)^2*a (ax+b)^3/2a=(ax+b)^2$

i know this is a round about way of solving this but i dont understand the "how to" tackle the question, if im blindly using the formulas. but even with this method, i cant seem to be able to solve all the basic questions with this method like

$x(x+2)^{(1/2)}$

my answer is

$\frac{(x^3+2x^2)^{3/2}}{18x^3}$

but actual answer is

can someone help me with where i miss the concept and how? thanks in advance

Answer 1

Here's some specific bits of advice.

1. You need to practice. Almost any calculus text will have 1000's of integral exercises. But a book like Schaum's Outline "Calculus" has lots and lots of worked out examples plus tons of exercises for you. It's cheap on Amazon.

2. A general principle of integration is that you try to change what is ugly into something prettier. And one of the ugliest things in mathematics is when exponentiation and addition try to work together. In your example you have $(x+2)^{1/2}$. The exponent doesn't distribute across the plus sign. To make that prettier, you might substitute $u = x+2$ which makes your integrand $(u-2)u^{1/2} = u^{3/2}-2u^{1/2}$. So look at your integrand and ask yourself "What do I wish was different?" Then see if you can cause that difference.

3. After you integrate, check your work by differentiating. If you've made a mistake, this will find it AND probably shed light on why it was a mistake.

Answer 2

Here is how I see this integral (and maybe you will also like thinking about it this way):

I want to find $\int (ax+b)^3\ dx$ but the answer is not immediate. On the other hand this has an obvious visible similarity to $\int x^3 \ dx$ which I already know to be $\frac{x^4}{4}+C$. One way to relate the known information to the unknown, is to say: Perhaps I should guess that $\int(ax+b)^3\ dx = \frac{(ax+b)^4}{4}$. It's probably wrong, and if you differentiate $\frac{(ax+b)^4}{4}$ you see that it is wrong -- but it's also not far from correct. And you can fix it with a factor.

However, even that solution method is unsatisfying, because you will eventually find more complicated integrals which you just can't analyze this way. A better way to connect the known and unknown information, is to try to turn this into a strategy which holds for an entire class of integrals. If you approach it this way, it will turn out that the solution method will help you to compute $\int \sin^{10}x\cos^2x \ dx$, for example.

So what we would like to do is take $\int (ax+b)^3 \ dx$ and see if there is some way to justify merely "blotting out" the complicated $ax+b$ and just turn this into a simple $u=ax+b$. Then using knowledge of the integral $\int u^3\ du$, can we find the original integral?

Well, the mere substitution is not sufficient. Of course you can rename any object to be anything you want, so there is no problem writing $u=ax+b$ and then saying that the integral we want is $\int u^3\ dx$. The problem is the $dx$. We do not have methods of integrating a variable $u$, when the variable of integration is something else, like $x$. How can we fix this?

The answer is to try to exchange $dx$ for something else which is in terms of $du$. Indeed we can do this by thinking about the derivative.

$$ \frac{du}{dx} = \frac{d}{dx}(ax+b) = a $$

Therefore $\frac{du}{dx} = a$ and then we can rearrange the differential parts as $dx = \frac{du}{a}$. Now we substitute this into our integral and find

$$\int u^3 \ dx = \int u^3 \ \frac{du}{a}$$

and it is easy to use our earlier methods to see that this is equal to $\frac{1}{a}\cdot\frac{u^4}{4}+C$. And then this is equal to

$$\frac{(ax+b)^4}{4a}+C$$

and we're done! And not only are we done, but we actually now have an entire strategy (often called "$u$-substitution") which works not just for this or a small family of problems -- but in fact for a VERY large class of valuable problems!

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Let's use the method to solve $\int x(x+2)^{1/2}\ dx$. We now have a method whereby we can take a troublesome part of the integral, cover it up with a $u$, and then integrate in terms of $u$. The troublesome part here is $(x+2)^{1/2}$. It's not entirely clear if we should cover up the whole thing and say $u=(x+2)^{1/2}$, or if we should just cover up $x+2$. Use this technique enough times and you tend to develop instincts for this kind of thing, and I will just cover up $x+2$.

So let $u=x+2$ so that our integral is now $\int xu^{1/2}\ dx$. This time not only do we need to exchange $dx$ for something with $du$ -- we also need to get rid of $x$. We can do both fairly easily. Since $u=x+2$ then $x=u-2$. Also since $\frac{du}{dx} = 1$ then $du=dx$. Then

$$\begin{aligned} \int x(x+2)^{1/2}\ dx &= \int (u-2)u^{1/2} \ du \\ &= \int (u^{3/2}-2u^{1/2})\ du \\ &= \int u^{3/2}\ du - 2\int u^{1/2} \ du \\ &= \frac{u^{5/2}}{5/2}-2\cdot\frac{u^{3/2}}{3/2} + C \\ &= \frac{(x+2)^{5/2}}{5/2}-2\frac{(x+2)^{3/2}}{3/2}+C \end{aligned}$$