A query in integration using method of substitution

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The discussion focuses on the method of substitution for solving integrals of the form ∫(lx + m)/(ax² + bx + c) dx. A participant expresses confusion about how to reach a specific part of the solution, particularly regarding the integral k₁∫(2ax + b)/(ax² + bx + c) dx, noting that the book simplifies it in one step. Another contributor advises against relying on online integral calculators, emphasizing the importance of practicing by hand to prepare for exams. They explain that the integral involves a linear function divided by a quadratic, which can often be approached by recognizing the relationship between the numerator and the derivative of the denominator. Ultimately, the discussion highlights the balance between using resources for assistance and developing problem-solving skills independently.
Wrichik Basu
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Homework Statement

:[/B]

I was learning the use of standard forms in method of substitution in solving integration. My book has given this method for solving integrals of the type ##\int \frac{lx +m}{ax^2+bx+c} dx##:

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As an example, the book gives this one:

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Homework Equations

:[/B]

The Attempt at a Solution



How does one solve to get the circled part? I can understand that the book is separating the two parts of the numerator. I also know how to solve the second part, but how does one reach the circled part?
 

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For ##k_1\int \frac {2ax+b} {ax^2+bx+c}\, dx## you need another substitution.
 
Arman777 said:
For ##k_1\int \frac {2ax+b} {ax^2+bx+c}\, dx## you need another substitution.
Even I was thinking that. How do I solve that part: ##k_1 \int \frac{2ax +b}{ax^2+bx+c} dx##? As per the book, I do not require another substitution. So, how should I proceed with that part?
 
Searching the Internet for something called an "integral calculator", I found one, and it showed me the steps properly:

20180225_112639.png


This solves my problem. The book had done it in one step, which is why I was facing the problem.
 

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It's nice but it would be better If you work by yourself and find the solution. Anyways yes, that's how you do it.
 
Wrichik Basu said:
Searching the Internet for something called an "integral calculator", I found one, and it showed me the steps properly:

View attachment 221031

This solves my problem. The book had done it in one step, which is why I was facing the problem.

When you are a beginner, just starting to learn the subject, you should avoid such on-line tools (except, possibly to check your work). You will never figure out how to do integrals without doing lots of examples by hand and without assistance from computer-aided integration tools. Think of it this way: what would you do on an exam, where you have no access to such facilities?
 
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Ray Vickson said:
When you are a beginner, just starting to learn the subject, you should avoid such on-line tools (except, possibly to check your work). You will never figure out how to do integrals without doing lots of examples by hand and without assistance from computer-aided integration tools. Think of it this way: what would you do on an exam, where you have no access to such facilities?
I solve problems from at least two books before proceeding to a new topic. Help materials like online calculators help in self study and nothing else, especially when you're stuck at a problem you just can't solve.
 
The expression to be integrated is a linear function divided by quadratic. A linear function is the derivative of a quadratic. So if you were lucky the numerator would be the derivative of the denominator, i.e. the expression to be integrated wrt x would be ##\dfrac {f'\left( x\right) }{f\left( x\right) }## which I presume you know how to do. In this case we are as usual not so lucky, but we can make it into a linear part which is part which is derivative of the denominator (just hammering the constants) with some other constant leftover – that second part then being a constant divided by a quadratic which we know how to integrate. Well actually many people would think that second part is the most difficult one.
 
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